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DESCRIPTIVE   GEOMETRY 


THE  MACMILLAN   COMPANY 

NEW  YORK    •    BOSTON    -    CHICAGO  •    DALLAS 
ATLANTA   •    SAN    FRANCISCO 

MACMILLAN  &  CO.,  Limited 

LONDON  •  BOMBAY  •  CALCUTTA 
MELBOURNE 

THE  MACMILLAN  CO.  OF  CANADA,  Ltd. 

TORONTO 


DESCRIPTIVE   GEOMETRY 


BY 

ERVIN   KENISON 

ASSOCIATE    PROFESSOR    OF    DRAWING    AND    DESCRIPTIVE    GEOMETRY 
MASSACHUSETTS    INSTITUTE    OF    TECHNOLOGY 

AND 

HARRY   CYRUS   BRADLEY 

A.SSI8TAMT    PROFESSOR    OF    DRAWING    AND    DESCRIPTIVE    GEOMETRY 
MASSACHUSETTS    INSTITUTE    OF    TECHNOLOGY 


Ncto  ©ork 

THE   MACMILLAN   COMPANY 

1918 

All  rights  reserved 


Copyright,  1916  and  1917, 
By  THE   MACMILLAX   COMPANY. 


Set  up  and  elcctrotyped.     Preliminary  edition  published 
October,  1916.     Complete  edition,  October,  1917. 


ICormooU  ^rcss 

J.  S.  Cushing  Co.  —  Berwick  &  Smith  Co. 

Norwood,  Mass.,  U.S.A. 


PREFACE 

This  book  represents  a  teaching  experience  of  more  than 
twenty  years  on  the  part  of  both  of  the  authors  at  the  Massa- 
chusetts Institute  of  Technology.  In  presenting  such  a  book 
for  the  consideration  of  other  institutions  and  of  the  public 
at  large,  it  is  well  to  note  the  relation  of  this  course  in  de- 
scriptive geometry  to  the  general  study  scheme,  in  order  that 
the  point  of  view  of  the  authors  may  be  better  understood. 

In  any  professional  drawing  of  an  engineering  or  architec- 
tural nature,  especially  if  the  element  of  design  occurs,  the 
draftsman  or  designer  must  see  clearly  the  conditions  in  space. 
This  has  been  well  expressed  by  saying  that  he  must  be  able 
to  think  in  three  dimensions.  Such  an  ability  is  natural  to 
but  few.  Fortunately,  however,  it  is  a  power  which  can  be 
acquired,  more  or  less  readily,  by  the  great  majority.  This 
power  can  be  gained,  and  it  usually  is  by  the  so-called  "  prac- 
tical "  draftsman,  by  the  simple  process  of  making  and  re- 
making working  drawings.  But  the  authors  believe  that  the 
same  power  can  be  acquired  much  more  rapidly,  and  when 
acquired  can  be  more  forcefully  and  efficiently  applied  in  de- 
signing, through  the  study  of  descriptive  geometry. 

The  point  of  view  in  this  text  is  therefore  that  of  the  drafts- 
man. Mathematical  formula;  and  analytic  computations  have 
been  almost  entirely  suppressed.  It  has  been  found  that  the 
students  readily  apply  their  knowledge  of  the  theoretical 
mathematics  to  a  finished  drawing.  For  example,  they  make 
trigonometric  computations  from  drawings  with  considerable 
facility.  On  the  other  hand,  in  applying  even  the  simplesl 
principles  of  solid  geometry  during  the  construction  of  a 
drawing  the  student,  is  often  anything  bul  facile.  The  method 
of  attack  throughout  this  book  is  intended  to  be  that  which 

v 


vi  PREFACE 

shall  most  clearly  present  the  actual  conditions  in  space. 
Wherever  experience  has  shown  that  a  simple  plan  and  ele- 
vation are  not  amply  sufficient  for  this  purpose,  additional 
views  or  projections  have  been  introduced  freely,  correspond- 
ing to  the  actual  drafting  practice  of  making  as  many  side 
views  or  cross  sections  as  may  be  needed. 

In  the  matter  of  arranging  the  views  in  a  practical  drawing, 
there  has  been,  still  is,  and  probably  always  will  be,  contro- 
versy. Architects,  who  must  embellish  their  drawings  with 
shades  and  shadows,  as  well  as  some  engineers,  prefer  to  place 
the  plan  below  the  elevation,  according  to  the  so-called  first 
quadrant  projection.  Entire  text-books,  noticeably  the  older 
ones,  have  been  prepared  with  all  the  objects  and  all  the 
theory  studied  in  the  first  quadrant.  Other  engineers,  me- 
chanical especially,  insist  that  the  only  proper  placing  of  the 
views  is  to  have  the  plan  above  the  elevation,  according  to 
the  method  of  the  third  quadrant.  In  consequence,  some 
recent  text-books  have  been  issued  with  all  the  work  in  the 
third  quadrant.  The  authors  have  no  desire  to  take  sides  in 
this  controversy,  but  believe  in  giving  the  student  practice 
in  both  of  these  methods.  Hence  some  of  the  problems  are 
studied  in  the  first  quadrant,  others  in  the  third.  It  has  been 
found  that  the  student  has  no  greater  difficulty  in  working  in 
one  quadrant  than  in  the  other,  while  considerable  additional 
power  is  gained,  and  a  clearer  insight  into  space  in  general  is 
obtained,  by  not  restricting  the  work  to  a  particular  corner 
of  space. 

A  requirement  of  practical  drafting  is  that  the  construction 
shall  be  confined  to  the  limits  of  the  drawing  board.  There 
is  always  a  limit  beyond  which  points  cannot  be  made  avail- 
able. A  similar  requirement,  namely,  that  all  the  points  used 
in  the  construction  must  be  within  the  limits  of  the  figure,  is 
rigidly  insisted  upon  in  the  authors'  classes.  This  frequently 
involves  an  auxiliary  construction  not  embodied  in  the  gen- 
eral theory  of  the  problem.  Numerous  auxiliary  constructions 
of  this  kind  are  given  throughout  the  book.  The  additional 
grasp  of  the  subject  obtained  by  being  obliged  thus  to  force 


PREFACE  vil 

his  way  through  whatever  difficulties  may  arise,  has  been 
found  of  great  value  to  the  student  in  his  subsequent  work. 

The  amount  of  ground  covered  by  this  book  is  that  which 
is  considered  sufficient  to  enable  the  student  to  begin  the 
study  of  the  technical  drawings  of  any  line  of  engineering  or 
architecture.  It  is  not  intended  to  be  a  complete  treatise  on 
descriptive  geometry.  Detailed  exposition  of  such  branches 
as  shades  and  shadows,  perspective,  stereographic  projection, 
axonometry,  the  solution  of  spherical  triangles,  etc.,  will  not 
be  found.  The  student  is  prepared  by  the  present  text,  how- 
ever, to  take  up  any  of  these  subjects. 

The  notation  adopted  in  this  book  is  a  modification  of  that 
first  suggested  and  used  by  Professor  William  Watson  of  the 
Institute  of  Technology  (around  1870).  The  modifications 
have  been  made  from  time  to  time  by  the  authors  in  the 
course  of  their  teaching  experience.  The  authors  have  exam- 
ined many  books  ;  and  when  opportunity  offered,  they  have 
questioned  students  coming  from  other  institutions  in  their 
search  for  a  clear  system  of  notation.  While  the  system  here 
given  is  not  above  criticism,  the  authors  are  firmly  of  the 
opinion  that  it  is  the  best  that  has  ever  come  to  their  attention. 

Although  this  book  has  been  prepared  with  the  needs  of  one 
particular  institution  in  mind,  it  is  hoped  that  it  will  not  be 
found  wanting  in  general  interest. 

ERVIN    KENISON, 

HARRY   CYRUS   BRADLEY. 

Massachusetts  Ifstiti  n    of  Technology, 

September,  1917. 


CONTENTS 

PAGE 

Introduction 1 

Descriptive  Geometry  —  Definition  —  Visualization  — 
Practical  Applications. 

I.  Elementary  Principles  :   the  Coordinate  Planes       .         3 

Orthographic  Projection  —  Orthographic  View  — 
Projection  on  a  Single  Plane  —  Plane  of  Projection  — 
Choice  of  the  Planes  of  Projection  —  The  Horizontal 
and  Vertical  Coordinate  Planes  —  Names  of  the  Views 
—  Plan  —  Elevation  —  Position  of  the  Observer  —  The 
Ground  Line  —  The  Four  Quadrants  —  Relation  be- 
tween an  Object  and  Its  Projections  —  Relation  between 
Two  Projections  of  an  Object  —  Notation  —  Actual 
Projections. 

II.  Projections  of  the  Point  and  of  Simple  Solids       .       10 

Notation  of  the  Point  —  Projections  of  Isolated  Points 
in  All  Four  Quadrants  —  Projectors  —  Visualizing  the 
Quadrants  —  Uses  of  the  Quadrants  —  Distances  from 
H  and  V  —  Special  Positions  of  the  Point  —  Projections 
of  Simple  Solids  —  Visibility  of  Solid  Objects  —  Illus- 
trative Projections  of  the  Prism,  Pyramid,  Cylinder, 
Cone,  Sphere,  Torus. 

III.  Represf;ntation  of  the  Straight  Line:   Traces      .       19 
The  Straight  Line  —  Its  Projections  —  Notation  —  Vis- 
ualization —  Slope  —  Special  Positions  —  The  Profile 
Line  —  A  Point  Lying  in  a  Line  —  Traces  of  a  Line. 

Problem  1.     To  find  the  traces  of  a  straight  line  .         .       24 

Problem  2.     Given  the  two  traces  of  a  straight  line,  to  find 

ils  projections  ........       25 

IV.  Simple  Shadows     ........       26 

Shadows  —  Shadow  of  a  Point  —  Opaque  and  Trans- 
parent Planes  —  Shadow  of  a  Line  —  Of  Surfaces  —  Of 
a  Convex  Solid. 


x  CONTEXTS 

PAGE 

V.  Representation  of  the  Plane      .....       29 

The  Plane  —  Its  Representation  —  Traces  —  Notation 

—  Visualization  —  Slope  —  Special  Positions  —  Edge 
View. 

VI.  The  Profile  Plane  of  Projection     ....       33 

The  Profile  Plane  —  A  Profile  Projection  —  Relation  of 
Profile  Projection  to  H-  and  F-Projections  —  Profile 
Projections  in  All  Four  Quadrants  —  Profile  Projection  of 
the  Straight  Line  —  Problems  on  the  Profile  Line  —  Its 
True  Length  —  Its  Traces  on  H  and  V  —  Projections 
of  a  Point  Lying  in  the  Line  —  Profile  Trace  of  a  Line 

—  Of  a  Plane  —  Left-Side  Views  —  Conventional 
Placing  of  Views  —  Second  Method  of  Obtaining  Pro- 
file Projections. 

VII.  Secondary  Planes  of  Projection      ....       46 
Secondary  Planes  of  Projection  —  The  Profile  Plane  as 

a  Secondary  Plane  —  Secondary  Ground  Lines  —  Prin- 
ciples of  Secondary  Projection  —  Simplification  of  Prob- 
lems —  End  View  of  a  Line  —  Edge  View  of  a  Plane 

—  Secondary  Projections  of  Solids  —  The  Two  7-Pro- 
jections  Compared  —  Oblique  Secondary  Planes. 


VIII.    Revolution    of    a    Point  :     True    Length    of    a 

Straight  Line  :  Applications  .....  52 
Revolution  of  a  Point  about  a  Straight  Line  —  Axis  of 
Revolution  Perpendicular  to  H  or  V  —  Axis  Lying  in 
a  Given  Plane  —  True  Length  of  a  Line  —  Angles 
which  a  Line  Makes  with  H  and  V  —  Converse  Prob- 
lems. 

Problem  3.     To    find    the    true    length    of    a    line    (Two 

methods) 54,  56 

Problem  4.  To  find  the  projections  of  a  line  of  definite 
length,  when  its  slope,  the  angle  which  it  makes  with 
one  coordinate  plane,  and  the  direction  of  its  projection 
on  that  plane,  are  known        .......       58 

Problem  5.     To  find  the  projections  of  a  line  making  given 

angles  with  H  and  V 60 


CONTENTS  Xi 

PAGE 

IX.  Some  Simple  Intersections  :   Developments      .         .       62 

Simple  Intersections  of  Solids  of  Revolution  —  Visibility 
of  the  Intersection  —  A  Plane  Perpendicular  to  H  or  V 
Intersecting  a  Cone,  Sphere,  Torus,  Hyperbolic  Spindle 

—  Developments  —  Working  Method  for  Finding  the 
True  Length  of  a  Line  —  Development  of  a  Solid  with 
Plane  Faces  —  Projection  of  a  Prism  Whose  Long  Edges 
Make  Given  Angles  with  H  and  V. 

X.  Lines  in  a  Plane  :     Parallel  Lines  and  Planes       .       71 

Intersecting  and  Parallel  Lines  —  Parallel  Lines  in  Space 

—  Intersecting  Lines  in  Space  —  Test  for  Intersecting 
Lines  —  A  Line  in  a  Plane  —  A  Plane  Containing  a 
Given  Line  —  Principal  Lines  of  a  Plane  —  To  Pro- 
ject the  Principal  Lines  —  A  Plane  Containing  a  Line 
Parallel  to  H  or  V  —  Parallel  Planes  —  A  Line  Parallel 
to  a  Plane  —  A  Plane  Parallel  to  a  Line  —  The  Plane 
Determined  by  Two  Intersecting  Lines,  Two  Parallel 
Lines,  a  Line  and  a  Point  not  on  the  Line,  Three  Points 
not  in  the  Same  Straight  Line  —  Planes  Parallel  to  Lines 
or  to  Other  Planes  —  Use  of  Auxiliary  Lines  in  Finding 
the  Traces  of  Planes. 

Problem  6.     To  find  the  plane  which  contains  two  given 

intersecting  or  parallel  lines    ......       79 

Corollary  I.     To  find  the  plane  which  contains  a  given 

line  and  a  given  point     .......       82 

Corollary   II.     To  find  the  plane  which  contains  three 

given  points  not  in  the  same  straight  line  ...  82 
Problem  7.     To  find  the  plane  which  contains  a  given  line 

and  is  parallel  to  a  second  given  line  ....  84 
Problem  8.     To   find    the   plane   which   contains   a  given 

point  and  is  parallel  to  each  of  two  given  lines  .  .  80 
Problem  9.     To  find  the  plane  which  contains  a  given  point 

and  is  parallel  to  a  given  plane       .....       86 

XI.    Perpendicular  Lines  and  Planes       ....       91 
Perpendicular  Lines  —  Perpendicular  Lines  Whose  Pro- 
jections are  Perpendicular  —  A  Line  Perpendicular  to 
a  Plane  —  Test  for  Perpendicularity  of  a  Line  and  Plane 
—  Perpendicular  Planes  —  Lines  of  Maximum  Inclina- 


xii  CONTENTS 


PAGE 


tion  to   H  and   V  —  Planes  Perpendicular  to  a  Given 

Line  or  Plane. 
Problem  10.     To   find   the  plane  which  contains  a  given 

point  and  is  perpendicular  to  a  given  line       ...       95 
Problem  11.     To  find  the  plane  which  contains  a  given  line 

and  is  perpendicular  to  a  given  plane     ....       96 

XII.  Intersection  of  Planes  and  of  Lines  and  Planes  : 
Applications  ........       98 

Intersecting  Planes  —  Line  of  Intersection  of  Two  Planes 

—  Intersection  of  a  Line  and  a  Plane  —  Shortest  Dis- 
tance from  a  Point  to  a  Plane  —  Projection  of  a  Point  or 
Line  on  a  Plane. 

Problem  12.     To  find  the  line  of  intersection  of  two  planes       98 

Problem  13.     To  find  the  point  in  which  a  straight  fine 

intersects  a  plane    ........     103 

Problem  14.     To  find  the  shortest  distance  from  a  point  to 

a  plane   ..........     105 

Problem  15.     To  project  a  line  on  an  oblique  plane     .         .     106 

XIII.  Intersection   of   Planes  and   Solids  Bounded  by 
Plane  Faces         ........     108 

A  Plane  Determined  by  Two  Lines  —  Intersection  of  a 
Line  with  a  Plane  Determined  by  Two  Lines  —  Inter- 
section of  Two  Limited  Plane  Surfaces  —  Visibility  — 
Intersection  of  a  Plane  and  a  Pyramid  —  Of  Two  Solids 
Bounded  by  Plane  Faces  —  Of  a  Sphere  and  a  Prism  — 

Of  a  Sphere  and  a  Right  Cylinder  —  Of  a  Cylinder  and 
a  Torus. 

XIV.  Problems  Involving  the  Revolution  of  Planes  .     125 
A  Line  Lying  in  a  Given  Plane  —  A  Point  in  a  Plane  — 
Revolution  of  a  Plane  about  an  Axis  Perpendicular  to 

H  or   V  —  Distance   between  Two  Parallel  Planes  — 
Angles  between  an  Oblique  Plane  and  the  Coordinate 
Planes  —  Planes  making  Given  Angles  with  H  and  V. 
Problem  16.     Given  one  projection  of  a  line  lying  in  a  plane, 

to  find  the  other  projection     ......     125 

Corollary  I.     Given  one  projection  of  a  point  lying  in  a 
plane,  to  find  the  other  projection  ....     128 


CONTEXTS  xiii 

PAGE 

Corollary  II.  To  find  the  second  projection  of  a  line 
lying  in  a  plane  when  the  general  solution  fails,  partially 

or  wholly 129 

Corollary  III.  To  find  a  line  of  maximum  inclination  of 
a  plane   ..........     129 

Problem  17.     To  find  the  perpendicular  distance  between 

two  parallel  planes  .......     131 

Corollary.  To  find  the  perpendicular  distance  from  a 
point  to  a  plane      ........     132 

Problem  18.     To  find  the  angles  which  an  oblique  plane 

makes  with  H  and  V      ......  132 

Problem  19.  Given  one  trace  of  a  plane,  and  the  angle 
which  the  plane  makes  with  either  H  or  V,  to  find  the 
other  trace  of  the  plane  ......     134 

Problem  20.     Given  the  angles  which  a  plane  makes  with  H 

and  V,  to  find  the  traces  of  the  plane    ....     135 

XV.   Other    Problems    Involving    the    Revolution    of 

Planes 139 

Revolution  of  a  Plane  about  One  of  Its  Traces  —  Re- 
volved Position  of  a  Point  in  a  Plane  —  Working  Rule  — 
Solution  of  Plane  Problems  by  Revolution  of  the  Plane 
—  Angle  between  Two  Intersecting  Lines  —  Angle 
between  a  Line  and  a  Plane  —  A  Rectilinear  Figure 
Lying  in  a  Plane 

Problem  21.     To  find  the  position  of  a  point  lying  in  a  plane, 
when  the  plane  is  revolved  into  //  or  V  about  the  corre- 
sponding trace        .         .         .         .         .         .         •     b  •     139 

Corollary.  To  find  the  revolved  position  of  a  line  lying  in 
a  plane,  when  the  plane  is  revolved  into  //  or  V  about 
the  corresponding  trace  ......     142 

Problem  22.     To  find  the  angle  between  two  intersecting 

lines 143 

Corollary.  To  find  (he  projections  of  the  bisector  of  the 
angle  between  two  intersecting  lines      ....     145 

Problem  23.     To  find  the  angle  between  ;i  line  and  a  plane     147 

Problem  24.  Given  one  complete  projection,  and  three 
points  iii  the  older  projection  of  a  plane  polygon,  to 
complete  the  projection  ......     142 


xiv  CONTENTS 

PAGE 

Corollary  I.  To  find  the  true  size  and  shape  of  a  plane 
figure      ..........     150 

Corollary  II.  To  find  the  projection  of  the  line  which  bi- 
sects one  of  the  interior  angles  of  a  plane  polygon   .         .     151 

XVI.  Miscellaneous  Problems  of  the  Line  and  Plane     152 
Distance  from  a  Point  to  a  Line  —  Distance  between  Two 
Lines  —  Angle  between  Two  Planes  —  Application. 

Problem  25.     To  find  the  shortest  (perpendicular)  distance 

from  a  point  to  a  line     .......     152 

Problem  26.  To  find  the  shortest  distance  between  two 
lines  not  in  the  same  plane,  and  the  projections  of  their 
common  perpendicular   .......     155 

Problem  27.     To  find  the  angle  between  two  planes    .         .     158 

Problem  28.     Given  the  horizontal  traces  of  two  planes, 
and  the  angle  each  plane  makes  with  H,  to  find  the  line 
of  intersection  of  the  two  planes     .         .         .         .         .162 
Corollary.     To  find  the  angle  between  the  given  planes     163 

XVII.  Counter-revolution  of  Planes      ....     164 
Counter-revolution  of  a  Point,  a  Line,  a  Plane  Figure  — 

A  Line  of  Given  Length  Perpendicular  to  a  Plane  — 
Projections  of  a  Prism  or  Pyramid  Whose  Axis  is  In- 
clined to  Both  H  and  V. 

Problem  29.     Given  the  position  of  a  point  lying  in  a  plane 
after  the  plane  has  been  revolved  into  H  or  V  about  the 
corresponding  trace,  to  find  the  projections  of  the  point     164 
Corollary.     To  counter-revolve  a  plane  poljTgon        .         .     166 

Problem  30.  At  a  given  point  in  a  plane,  to  draw  a  line 
which  shall  be  perpendicular  to  the  plane  and  of  given 
length 168 

XVIII.  Tangent    Lines    and    Planes  :     General    Prin- 
ciples   ..........     174 

Curves  —  Plane  Curves  —  Space  Curves  —  Projections 

of  Curves  —  Tangent  Lines  —  Tangents  to  Plane  Curves 
—  Tangent  Planes  to  Curved  Surfaces  —  Determina- 
tion by  Means  of  Tangent  Lines  —  Rectilinear  Ele- 
ments —  The  Normal  —  Determination  of  Tangent 
Planes  by  Means  of  the  Normal. 


CONTEXTS  XV 

XIX.  Tangent  Planes  to  Cones  and  Cylinders  .  .  178 
The  Cone  and  Cylinder  —  Definitions  —  Representation 
—  Projections  of  Cones  and  Cylinders  —  Projections  of 
a  Point  in  the  Surface. —  Tangent  Planes  to  Cones  and 
Cylinders  at  a  Given  Point  in  the  Surface  —  Through 
a  Given  Point  without  the  Surface  —  Parallel  to  a  Given 
Line. 

Problem  31.     To  pass  a  plane  tangent  to  a  cone  at  a  given 

point  in  the  surface         .......     184 

Problem  32.     To  pass  a  plane  tangent  to  a  cone  through  a 

given  point  without  the  surface.      (Two  results.)    .         .     189 

Problem  33.     To  pass  a  plane  tangent  to  a  cone  parallel  to 

a  given  line.      (Two  results.)  ......     192 

Problem  34.     To  pass  a  plane  tangent  to  a  cylinder  at  a 

given  point  in  the  surface       ......     194 

Problem  35.     To  pass  a  plane  tangent  to  a  cylinder  through 

a  given  point  without  the  surface.     (Two  results.)         .     196 

Problem  36.     To  pass  a  plane  tangent  to  a  cylinder  parallel 

to  a  given  line.     (Two  results.)       .....     198 


XX.    Tangent  Planes  to  Double  Curved  Surfaces  of 

Revolution  .........     201 

Double  Curved  Surfaces  of  Revolution  —  Represen- 
tation —  Meridians  —  Principal  Meridian  Plane  — 
Parallels  —  Projections  of  a  Point  in  the  Surface  — 
—  The  Sphere  —  The  Torus  —  Tangent  Planes  to 
Double  Curved  Surfaces  of  Revolution  at  a  Given  Point 
in  the  Surface  —  Tangent  Planes  Which  Contain  a  Given 
Line. 

Problem  37.     To  pass  a  plane  tangent  to  a  sphere  at  a  given 

point  in  the  surface         .......     206 

Problem  38.     To  pass  a  plane  tangent  to  a  double  curved 

surface  of  revolution  at  a  given  point  in  the  surface       .     207 

Problem  39.     To  pass  a  plane  tangent  to  a  sphere  through  a 

given  line  without  the  surface         .....     212 

Problem  40.  To  pass  a  plane  tangent  to  a  double  curved 
surface  of  revolution  through  a  given  line.  (Special 
cases.)     .         .         .         .         .         .         .         .         .         .216 


xvi  CONTENTS 

XXI.  The  Intersection  of  Curved  Surfaces  by  Planes  219 
Classification  of  Curved  Surfaces  —  Ruled  Surfaces  — 
Surfaces  of  Revolution  —  Other  Curved  Surfaces  —  In- 
tersection by  a  Plane  of  a  Ruled  Surface  —  Of  a  Sur- 
face of  Revolution  —  Of  Any  Curved  Surface  —  Method 
by  Means  of  Secondary  Projections  —  Visibility  of  the 
Intersection  —  A  Rectilinear  Tangent  to  the  Curve  of 
Intersection  —  Development  of  a  Curved  Surface. 


XXII.    Intersection  of  Planes  with  Cones  and  Cylin- 
ders       .  222 

Intersection  of  a  Cone  or  Cylinder  with  a  Plane  —  A 
Line  Tangent  to  the  Section  —  Development  of  the 
Curved  Surface  and  Tangent  Line  —  Intersection  of  a 
Pyramid  or  Prism  with  a  Plane. 

Problem  41.     To  find  the  intersection  of  a  cone  and  plane     222 

Problem  42.     To  find  the  intersection  of  the  frustum  of  a 

cone  and  a  plane    ........     232 

Problem  43.     To  find  the  intersection  of  a  cylinder  and  a 

plane      ..........     236 

Problem  44.     To  find  the  intersection  of  a  pyramid  or  prism 

and  a  plane    .........     244 


XXIII.    Intersection     of     Planes     with     Surfaces     of 

Revolution  .........     247 

Intersection  of  a  Surface  of  Revolution  and  a  Plane  — 
Points  Determined  by  Meridian  Planes  —  A  Line  Tan- 
gent to  the  Intersection  —  Number  of  Curves  in  the 
Case  of  a  Torus  Intersected  by  a  Plane. 
Problem  45.     To  find  the  intersection  of  a  double  curved 

surface  of  revolution  and  a  plane    .....     247 


XXIV.   The     Intersection     of     Curved     Surfaces     by 

Curved  Surfaces  .......     256 

Intersection  of  Two  Curved  Surfaces  —  The  Auxiliary 
Surfaces  —  Visibility     of     the     Intersection  —  A     Line 
Tangent  to  the  Curve  of  Intersection. 


CONTENTS  xvii 

XXV.  The  Intersection  of  Cones  and  Cylinders  with 

Each  Other 258 

Intersections  of  Cones  and  Cylinders  —  Of  Two  Cones, 
Two  Cylinders,  a  Cylinder  and  a  Cone  —  Auxiliary 
Planes  Which  Cut  Elements  from  Each  Surface  —  Num- 
ber of  Curves  —  General  Cases  —  Special  Cases  — 
Limiting  Planes  Doubly  Tangent  —  Parallel  Elements 

—  Infinite  Branches  —  Intersection  of  Cylinders  and 
Cones  of  Revolution. 

Problem  46.     To  find  the  intersection  of  two  cylinders        .  260 
Problem  47.     To  find  the  intersection  of  a  cylinder  and  a 

cone 270 

Problem  48.     To  find  the  intersection  of  two  cones     .         .  272 

XXVI.  The  Intersection  of  Various  Curved  Surfaces     278 
Intersection  of  Two  Curved  Surfaces  of  Revolution  — 

Of  a  Sphere  and  a  Cone  —  Of  a  Sphere  and  a  Cylinder 

—  Of  Any  Two  Curved  Surfaces. 

Problem  49.     To  find  the  intersection  of  any  two  surfaces  of 

revolution  whose  axes  intersect       .....     278 
Corollary.     To  find  the  intersection  of  two  surfaces  of  rev- 
olution whose  axes  are  parallel        .....     280 

Problem  50.     To  find  the  intersection  of  a  sphere  and  a  cone     282 

Problem  51.     To  find  the  intersection  of  a  sphere  and  a 

cylinder 284 

Problem  52.     To  find  the  intersection  of  any  two  curved  sur- 
faces.    (General  case.) 286 


DESCRIPTIVE  GEOMETRY 


INTRODUCTION 

1.  Descriptive  Geometry.  Descriptive  Geometry  is  an  ap- 
plied science  which  treats  of  the  graphical  representation  of 
lines,  planes,  surfaces,  and  solids,  and  of  the  solution  of  prob- 
lems concerning  size  and  relative  proportions.  Thus  it  lies  at 
the  foundation  of  all  architectural  and  mechanical  drafting. 

While  the  study  of  descriptive  geometry  does  not  require 
extended  mathematical  knowledge,  and  while  its  operations  are 
not  strictly  mathematical,  the  best  results  cannot  be  obtained 
without  some  acquaintance  with  the  principles  of  plane  and 
solid  geometry. 

Descriptive  geometry  differs  from  analytic  geometry  of  three 
dimensions  in  that  the  solutions  are  based,  not  on  algebraic 
equations,  but  on  drawings.  In  a  certain  sense,  this  greatly 
increases  the  scope  of  descriptive  geometry.  By  the  graphical 
processes,  objects  of  any  form  whatever,  no  matter  how  irregu- 
lar, or  whether  any  equations  for  them  exist  or  not,  may  be 
treated.  On  the  other  hand,  descriptive  geometry  resembles 
analytic  geometry  in  that  the  drawings,  like  the  equations,  are 
merely  representations  of  the  conditions  in  space. 

By  the  methods  of  descriptive  geometry  the  solution  of  any 
problem  involving  three  dimensions  consists  of  three  distinct 
processes,  as  follows : 

(1)  Representation  of  the  lines,  planes,  surfaces,  or  solids 
in  space  by  correspond ing  plane  figures. 

(2)  Solution  of  the  problem  by  the  use  of  the  plane  figures. 

(3)  Determination  of  the  relation  in  space  which  corresponds 
to  this  solution. 

B  1 


2  DESCRIPTIVE  GEOMETRY  [§  1 

In  order  that  these  processes  may  lead  to  a  successful  result, 
it  is  evident  that  it  must  be  possible  to  pass  without  ambiguity 
from  the  object  in  space  to  its  representations,  and  also  with- 
out ambiguity  from  the  representations  to  the  object  in  space 
again. 

2.  Visualization.  The  process  of  passing  from  the  repre- 
sentations to  the  object  in  space  is  a  purely  mental  one.  It 
is,  therefore,  apt  to  be  ignored  by  the  student  at  the  beginning 
of  his  course,  to  his  detriment  later  on.  The  process  is  called 
"  visualizing "  or  "  reading "  the  drawing,  and  is  absolutely 
essential  in  any  practical  application  of  descriptive  geometry. 
(See  §  3.)  Because  of  its  importance,  both  in  theory  and  in 
practice,  considerable  emphasis  will  be  laid  on  visualization 
in  the  present  text.  The  student  is  advised  to  accustom  him- 
self early  to  look  upon  his  drawings  as  representations  of 
conditions  in  space,  and  to  work  out  his  problem  by  consider- 
ing the  space  relations  involved,  rather  than  by  the  (plane) 
geometrical  relations  existing  in  the  drawing  itself. 

3.  Practical  Application  of  Descriptive  Geometry.  In  the  de- 
sign of  all  engineering  and  architectural  structures,  as  machines, 
bridges,  buildings,  etc.,  as  well  as  in  many  of  the  less  preten- 
tious mechanic  arts,  there  comes  a  time  when  the  forms  and 
arrangements  of  the  various  parts  must  be  considered.  The 
problem  then  becomes  one  which  is  solved,  either  wholly  or 
in  part,  by  the  graphical  methods  of  descriptive  geometry. 
Moreover,  the  drawings,  once  made,  take  the  place  of  written 
description,  and  form  the  language  by  means  of  which  the 
designer  conveys  his  ideas  to  the  builder  or  mechanic,  who, 
after  "  reading "  (visualizing)  them,  can  build  or  construct 
the  work. 

In  addition  to  the  usefulness  of  descriptive  geometry  in  its 
practical  applications,  it  is  the  conviction  of  the  authors  that 
its  fundamental  and  educational  value  lies  in  its  unique  power 
to  develop  the  mental  concept,  or  visualization,  the  importance 
of  which  has  already  been  emphasized. 


CHAPTER   I 
ELEMENTARY  PRINCIPLES  —  THE   COORDINATE   PLANES 

4.  Orthographic  Projection  or  Orthographic  View.  The  terms 
orthographic  projection  and  orthographic  view,  or  more  simply 
projection  and  view,  are  used  to  denote  a  drawing  which  repre- 
sents, in  accordance  with  a  certain  artificial  and  conventional 
system  of  vision  hereafter  described,  some  object,  surface, 
line,  point,  or  combination  of  these,  that  has  a  definite  position 
in  space.  Such  a  drawing  may  be  made  on  any  plane  surface, 
as,  for  example,  a  sheet  of  paper  or  a  blackboard. 

5.  Projection  on  a  Single  Plane.  In  Fig.  1,  let  Q  represent 
any  horizontal  plane  in  space,  and  A  a  square  right  prism  with 


>' 

2 

3 

> 

1 

1 
1 

7 
A 

1 

X 

7 

i 

/          5 

r~v:/ 

/a 

Fig.  1. 

its  bases  parallel  to  the  plane.  If  the  prism  be  looked  at 
from  above,  at  right  angles  to  the  plane,  the  arrows  represent- 
ing the  lines  of  sight,  the  square  top,  1-2-3-4,  will  be  the  only 
part  seen.  If  this  square  be  now  imagined  to  drop  vertically 
until  it  lies  in  the  plane  Q  in  the  position  5-6-7-8,  the  latter 
would  be  the  orthographic  projection  of  the  prism  on  the 
plane  Q.  The  plane  Q  is  called  a  plane  of  projection.  In  this 
case  the  projection  is  a  top  view  of  the  given  object. 

3 


4  DESCRIPTIVE  GEOMETRY  [I,  §  5 

In  looking  at  any  object  naturally,  it  is  a  familiar  fact  that 
the  lines  of  sight  all  converge  to  the  eye.  In  orthographic 
projection,  however,  this  is  not  the  case  ;  instead,  the  lines  of 
sight  are  all  assumed  to  be  parallel,  and  every  view  or  projec- 
tion is  made  on  this  basis.  On  account  of  this  difference,  the 
resulting  views  are  more  or  less  unlike  those  seen  with  the 
natural  eye.  With  natural  vision,  the  same  object,  placed  at 
varying  distances  from  the  eye,  appears  smaller  when  farther 
away,  while  in  orthographic  projection  the  size  of  the  view  or 
projection  is  not  affected  by  the  distance  of  the  object  from 
the  observer. 

Figures  2-10  show  pictorially  the  projections  of  objects  of 
simple  form.     Figures  2,  3,  4,  5,  6,  and  9  are  projections  on  a 


Fig.  2. 

horizontal  plane,  and  are  top  views.  Figures  7,  8,  and  10  are 
projections  on  a  vertical  plane,  and  are  front  views.  In  each 
case  the  plane  Q  is  a  plane  of  projection. 

From  these  figures  it  is  evident  that  one  view  of  an  object  is 
not  sufficient  to  determine  its  size  and  shape,  hence  two  or 
more  views  or  projections,  requiring  as  many  planes  of  pro- 
jection, must  be  used. 

6.    Choice  of  the  Planes  of  Projection  ;  the  Coordinate  Planes. 

The  simplest  and  most  advantageous  angle  between  the  planes 
is  a  right  angle. 

At  any  given  point  On  the  earth's  surface,  there  are  two 
natural  mutually  perpendicular  directions,  which  must  always 
be  considered  in  the  applied  arts  and  sciences :  the  horizontal 
or  level,  as  shown  by  the  surface  of  still  water,  and  the  vertical 
or  plumb,  as  shown  by  a  freely  falling  body. 


I,  §6] 


ELEMENTARY  PRINCIPLES 


Fig.  7. 


Fig.  6. 


Fig.  9. 


Fig.  10. 


6  DESCRIPTIVE  GEOMETRY  [I,  §  6 

The  planes  of  projection  are  chosen,  •  accordingly,  the  first 
horizontal,  the  second  perpendicular  to  the  first,  and  therefore 
vertical.  These  two  planes  are  known  as  the  horizontal  and 
vertical  coordinate  planes,  respectively,  and  will  be  designated 
by  the  letters  H  and  V. 

At  any  given  place  the  horizontal  plane  is  always  fixed  in 
direction,  since  all  horizontal  planes  are  parallel.  But  any 
plane  perpendicular  to  the  horizontal  is  vertical.  Hence  the 
direction  of  the  vertical  plane  is  to  some  extent  arbitrary. 

7.  Names  of  the  Views.  A  view  or  orthographic  projection 
made  on  a  horizontal  plane  of  projection,  H,  is  known  as  a  hori- 
zontal projection,  usually  abbreviated  to  JJ-projection.  This 
term  is  equivalent  to  the  expression  top  vieiv  which  was  used 
above,  and  which  may  still  be  employed.  In  practice,  however, 
it  is  usually  called  the  plan.  Thus,  the  terms  top  view,  plan, 
and  horizontal  projection  are  synonymous.  Similarly,  a  view 
made  on  a  vertical  plane  of  projection,  V,  is  called  a  vertical 
projection,  or  a  F-projection,  or  a  front  view,  or  an  elevation. 

8.  Position  of  the  Observer  with  Reference  to  the  Coordinate 
Planes.  When  making  a  vertical  projection  or  front  view,  the 
observer  is  in  a  position  squarely  facing  the  F-plane,  and 
either  above  or  below  H,  as  the  size  and  position  of  the  object 
may  require.  When  making  a  top  view  or  if-projection,  the 
position  is  as  if  the  observer  were  to  stand  on  or  above  the 
-fiT-plane,  facing  the  F-plane,  then  to  bend  forward  and  look 
vertically  down  upon  the  given  object  and  the  horizontal  plane. 
Except  in  rare  instances,  not  considered  here,  the  F-plane  is 
never  viewed  from  the  back  or  further  side,  nor  the  iT-plane 
from  underneath. 

9.  The  Ground  Line  and  the  Four  Quadrants.  Let  the  coordi- 
nate planes  be  chosen  as  in  Figs.  11  and  12.  The  line  of 
intersection,  OL,  of  these  planes  is  known  as  the  ground  line 
—  a  somewhat  unfortunate  name,  since  it  is  derived  from  but 
one  of  the  properties  of  the  line,  and  ignores  another  equally 
important  property,  as  will  be  seen  later.  The  ground  line 
divides  each  of  the  coordinate  planes  into  two  parts. 


I,  §  11] 


ELEMENTARY  PRINCIPLES 


Although  a  pictorial  representation  can  show  only  a  limited 
portion  of  each  plane,  the  coordinate  planes  are  supposed 
indefinite  in  extent.  Hence  they  divide  the  whole  of  space 
into  four  quadrants,  numbered  I,  II,  III,  and  IV,  as  follows  : 

Quadrant  I :  above  H  and  in  front  of  (or  on  the  near  side 
of)  V. 

Quadrant  II :  above  H  and  behind  (or  on  the  far  side 
of)  V. 

Quadrant  III :   below  H  and  behind  V. 

Quadrant  IV  :    below  H  and  in  front  of  V. 

10.  The  Relation  between  an  Object  and  Its  Projection.     In 

Fig.  1,  the  projection  5-6-7-8  of  the  top  of  the  prism  might  be 
obtained  by  extending  the  vertical  edges  of  the  prism  until  they 
intersect  the  plane  of  projection,  Q.  The  point  5  is  the  pro- 
jection of  point  1,  6  of  2,  and  so  on.  In  general,  the  (ortho- 
graphic) projection  of  a  point  on  any  plane  may  be  defined 
as  the  foot  of  a  perpendicular  from  the  point  to  the  plane. 

The  projection  of  any  object  is  composed  of  the  projection 
of  all  its  points,  found  in  the  same  manner. 

11.  The  Relation  between  Two  Projections  of  an  Object.  Let 
a  rectangular  card,  abed,  be  placed  as  shown  in  Fig.  11.  The 
card  is  in  the  first  quadrant,  above 
II,  in  front  of  and  parallel  to  V. 
The  long  edges  are  vertical,  the 
short  edges  parallel  to  H.  The 
projection  of  the  card  on  F~is  the 
equal  rectangle  avbvcvdv,  and  on  // 
the  straight  line  a*6*c*d*,  equal  in 
length  to  the  short  edge.  The 
perpendiculars  dd"  and  ddh  deter- 
mine a  plane  which  is  perpendicu- 
lar to  both  II and  V,  and  therefore 
bo  their  line  <>('  intersection,  GL. 
The  same  is  true  for  the  two  perpendiculars  from  every  other 
corner  of  the  card.  Hence  the  tivo  projections  of  any  jwint, 
together  with  the  point  itself,  must  always  be  in  the  same  plane 


Fig.  11. 


8 


DESCRIPTIVE  GEOMETRY 


[I,  §  11 


perpendicular  to  GL.     This  is  one  of  the  fundamental  relations 
of  orthographic  projection. 

12.  Notation.  In  this  book,  projections  on  the  horizontal 
coordinate  plane,  H,  will  be  denoted  by  the  small  letter,  h, 
attached  as  an  exponent  to  the  letter  or  character  which  denotes 
the  actual  point,  line,  or  object  projected.  Projections  on  the 
vertical  coordinate  plane,  V,  will  be  similarly  denoted  by  the 


Fig.  11  (repeated). 


Fig.  12. 


use  of  a  small  letter,  v,  as  an  exponent.  Thus,  in  Fig.  11,  point 
a  is  one  corner  of  the  card  in  space  ;  the  projection  of  a  on  H 
is  lettered  ah ;  of  a  on  V  is  called  a".  In  Fig.  12,  the  object,  A, 
a  rectangular  block  placed  in  the  third  quadrant,  is  projected 
on  H  as  Ah,  on  V  as  A".  Further  statements  in  regard  to 
the  notation  will  be  made  from  time  to  time,  when  necessary. 

13.  Monge's  Method.  The  use  of  two  mutually  perpen- 
dicular coordinate  planes,  as  in  Figs.  11  and  12,  enables  us  to 
represent  objects  of  three  dimensions  by  plane  figures.  It  is 
sometimes  called  Monge's  method,  after  the  originator. 
Drawings,  however,  are  not  made  ordinarily  on  two  drawing 
surfaces  at  right  angles  to  each  other.  A  single  drawing  surface 
(paper,  cloth,  board,  etc.)  is  taken  to  represent  both  H  and  V. 

14.  Projections.  In  the  pictorial  representations,  Figs.  1-12, 
the  planes  of  projections  have  been  shown  limited  in  extent, 
in  order  to  give  a  clear  idea  of  their  position.     Actually,  the 


I,  §  H] 


ELEMENTARY   PRINCIPLES 


9 


planes  are  unlimited  in  extent  (§  9),  so  that  in  making  pro- 
jections no  outlines  for  the  coordinate  planes  are  shown. 

The  projections  or  views  of  the  card  of  Fig.  11  are  given  in 
Fig.  13.     When  looking  toward  V,  the  horizontal  plane  will  be 


avi ib 


□: 


□* 


dh       ch 
Fig.  13. 


3  A" 


Fig.  14. 


seen  edgewise,  and  is  represented  by  GL ;  the  F~-projection  of 
the  card  will  be  the  rectangle  avbvcvdv,  as  in  Fig.  11.  Looking 
down  on  the  i/-plane,  the  P"-plane  will  be  seen  edgewise,  and 
is  usually  represented  by  the  same  line,  GL,  previously  used  as 
the  edge  view  of  H.  The  card  will  also  be  seen  edgewise,  and 
will  project  on  H  in  the  straight  line  ahdhbhch. 

The  projections,  plan  and  elevation,  of  the  rectangular  block 
of  Fig.  12  are  given  in  Fig.  14  at  Ah  and  Av  respectively. 

Figure  15  shows  the  projections  of  the  object  of  Fig.  3.  Fig- 
ure 16  shows  the  projections  of  the  hollow  ring  of  Fig.  2. 


^3 


Elev. 


Plan 


Plan 


]    Elev. 


Fig.  15 


Fig.  l(i 


Since  no  ground  line  is  given  in  Figs.  15  and  10,  the  quadrants 
in  which  these  objects  are  placed  are  not  known.  However, 
this  does  not  affect  the  size  or  the  shape  of  the  projections  (§  5). 


CHAPTER   II 

PROJECTIONS  OF  THE  POINT  AND  OF  SIMPLE  SOLIDS 

15.  Notation  of  the  Point.  An  isolated  point  in  space  will  be 
denoted,  in  general,  by  a  small  letter.  A  point  lying  in  a  line, 
or  in  the  boundary  of  a  surface  or  solid,  will  be  denoted  either 
by  a  small  letter  or  by  a  number,  according  to  convenience. 
The  projection  of  any  point  on  H  or  V  will  be  designated  by  the 

suitable  use  of  the  small  letters  h 
and  v  as  exponents  (§  12). 

16.  Projecting  Isolated  Points. 
I  Let  a,  Fig.  17,  be  a  point  situated 
1       in  Quadrant   I.     By  inspection   of 

^°h >v»j,iv\     the  figure,  it  is  seen  that  this  point 

projects  on  //  in  front  of  GL,  and 

on   V  above    GL.     Let   the  points 

b,  c,  cl,  be  placed  in  Quadrants  II, 

III,  IV,  respectively.     Visualizing  in  a  similar  manner,  and 

using  the  notation  of  §  15,  we  may  make  the  following  table : 

Quadrant      I :  ah  lies  in  //  in  front  of  GL. 

av  lies  in  V  above  GL. 
Quadrant    II :  bh  lies  in  H  behind  GL. 

bv  lies  in  V  above  GL. 
Quadrant  III :  ch  lies  in  H  behind  GL. 

c*  lies  in  V  below  GL. 
Quadrant  IV :  dh  lies  in  H  in  front  of  GL. 

dv  lies  in  V  below  GL. 

By  the  method  of  the  single  ground  line,  Fig.  13,  the  drawing 
is  divided  into  two  parts  by  the  line  GL.  The  upper  part 
represents  both  V  above  GL  and  H  behind  GL,  the  lower  part 
represents  both  H  in  front  of  GL  and  V  below  GL.  In  this 
method,  the  lines  joining  ah  and  a",  bh  and  bv,  etc.,  must  be  per- 
pendicular to  GL.    The  resulting  projections  are  given  in  Fig.  18. 

10 


II,  §  18]  POINTS  AND  SIMPLE  SOLIDS  11 

17.  Projectors.  The  line  aa"  or  bbv,  Fig.  11,  which  projects 
the  point  on  the  plane  of  projection,  is  called  a  projector.  A 
line  like  avah,  Fig.  13  or  Fig.  18,  which  connects  the  two  pro- 
jections of  a  point  in  the  drawing,  also  is  called  a  projector. 
These  two  uses  of  the  word  rarely  cause  confusion ;  if  necessary 

1  II  III  IV 


ib* 


i 
I 
I 


- 1 1 

•  l    h 

i 
-t-d 


Fig.  18. 

to  distinguish,  however,  the  terms  space  projector,  in  space,  and 
ruled  projector,  in  the  drawing,  may  be  employed. 

18.  Visualizing  the  Quadrants.  The  propositions  of  §  16  can 
be  reversed,  and  the  quadrant  in  space  in  which  any  point  lies 
can  be  told  at  once  from  its  projections.  To  do  this  merely  by 
memorizing  the  relations  of  the  two  projections  to  the  ground 
line  is  not  visualizing  in  any  sense,  and  is  of  little  or  no  value. 

As  the  first  step  in  visualizing,  let  us  inquire  which  side  of 
each  of  the  coordinate  planes  is  represented  by  the  drawing 
surface.  Regardless  of  the  position  of  the  point  or  object  rep- 
resented, this  is  always  the  side  of  the  plane  which  is  nearer 
to  an  observer  placed  in  the  first  quadrant.  The  same  idea  is 
expressed  by  saying  that  the  //-plane  is  always  viewed  from 
above,  and  the  F-plane  from  in  front  (§  8). 

Since  the  point  may  be  on  either  side  of  the  coordinate  plane 
while  the  drawing  is  always  viewed  from  the  same  side,  it 
follows  that  it  maybe  necessary  to  visualize  the  point  in  either 
of  two  ways :  (1)  nearer  to  the  observer  than  its  projection, 
that  is,  above  or  in  front  of  the  drawing ;  (2)  beyond  its  pro- 
jection, that  is,  below  or  behind  the  drawing  and  seen  as  if 
looking  through  a  transparent  surface. 


12 


DESCRIPTIVE  GEOMETRY 


[II,  §  19 


19.  Uses  of  the  Quadrants.  In  Figs.  19-22  an  isolated  point 
is  shown  placed,  successively,  in  the  four  quadrants.  The  re- 
marks and  deductions  apply  equally  well  to  any  object  lying 
wholly  within  the  quadrant  considered. 

First  Quadrant  (Fig.  19).  The  point  is  on  the  nearer  side 
of  the  drawing  in  each  projection.     This  quadrant  is   much 


Fig.  19. 
used  in  architectural  work  and  occasionally  in  engineering 
work.  On  account  of  the  comparative  ease  with  which  aids 
to  visualization  may  be  placed  over,  rather  than  under,  the 
drawing,  this  quadrant  is  very  generally  used  in  the  first 
presentation  of  a  problem  in  descriptive  geometry. 

Second  Quadrant  (Fig.  20).     The  point  is  on  the  nearer 


|bh 


Fig.  20. 


side  of  the  77-projection,  but  on  the  farther  side  of  the  F-pro- 
jection.  The  principal  use  of  this  quadrant  is  in  the  subject 
of  perspective. 

Third  Quadrant  (Fig.   21).     The  point  lies  beyond   the 
plane  of  the  drawing  in  each  projection,  and  must  be  imagined 


II,  §  20] 


POINTS  AND  SIMPLE  SOLIDS 


13 


as  if  seen  through  transparent  planes.     This  quadrant  is  used 
more  frequently  in  practical  drafting  than  any  of  the  others.  . 

J 


Yd 

~!ch 

W\ 

G 

1 
1 

! 
I 

di- 

k_lc 

L 

\ 

1 

This   is   analogous   to   the 


id* 
+d 


Fig.  21. 

Fourth  Quadrant  (Fig.   22). 
second  quadrant,  with 
H  and    V  reversed. 
It    is    little   used    in 
practice. 

20.  Views  of  a 
Prism.  The  plan  and 
elevation  of  a  hex- 
agonal right  prism 
placed  in  the  third 
quadrant  are  given  in 
Pig.  23.  The  prism 
is  in  a  vertical  position,  its  upper  base  in  the  if-plane.  Visualize 
the  solid  in  space.     The  distance  y,  the  length  of  the  prism, 

is  the  height  of  one  end  above  the 
other.  The  distance  x  is  the  dis- 
tance between  the  parallel  sides,  or 
the  thickness  of  the  prism. 

This  illustrates  one  of  the  funda- 
mental principles  of  orthographic 
projection :  an  elevation,  or  V-pro- 
jection,  shows  heights,  or  distances  up 
and  down  ;  a  plan^  or  1 1- projection, 
Fig.  23.  shows  distances  from  front  to  back. 


Fig.  22. 


1   1 

G 

I- 

t 

1 

y 

1 

Atf 

1 
L. 

X 

.-J 


14  DESCRIPTIVE  GEOMETRY  [II,  §  21 

21.  Distances  of  a  Point  from  H  and  V.  Let  it  be  required  to 
visualize  the  point  m,  Fig.  24.  First,  suppose  the  drawing 
paper  to  represent  H;  then  m*  is  ignored,  while  GL  becomes 
the  //"-projection  of  the  V  coordinate  plane.     Hence  m*  shows 

that  the  point  m  in  space  is  not  only  in 
~J  front  of  V,  but  also  at  the  distance  x  from  V. 

Now  suppose  the  drawing  paper  to  repre- 
sent V;  then  m*  is  ignored,  GL  becomes 
the  F-projection  of  the  incoordinate  plane, 
and  m"  shows  that  the  point  m  is  at  the 
j    v  distance  y  below  H. 

FlG  ^  Conversely,  if  the  distances  of  the  point 

in  space  from  H  and  V  are  given,  the  pro- 
jections can  be  located  at  the  given  distances  from  GL. 
Which  projection  is  determined  by  which  distance? 

22.  Special  Positions  of  the  Point.     In  Fig.  25, 
the  point  e  lies  in  H,  and  in  front  of  V; 

the  point  /  lies  in  H,  and  behind  V; 

i  r 


+en 


Fig.  25. 


the  point  g  lies  in  V  and  above  H; 

the  point  j  lies  in  both  H  and  F,  that  is,  in  the  ground  line. 

Visualize  each  of  these  points. 

23.  Projections  of  Simple  Solids.  Solids  bounded  by  plane 
faces,  such  as  prisms,  pyramids,  wedges,  frustums,  etc.,  are 
projected  by  drawing  the  projections  of  the  straight  lines 
which  form  the  edges  of  the  solid.  Only  simple  positions  of 
these  solids  can  be  considered  at  present. 


II,  §  25] 


POINTS  AND  SIMPLE  SOLIDS 


15 


24.  Visibility  of  Solid  Objects.  In  viewing  a  solid,  no  matter 
from  what  point  of  view,  only  a  portion  of  its  surface  is  visible, 
while  the  rest  is  invisible.  In  any  projection  of  a  solid,  there- 
fore, there  are  usually  both  full  and  dotted  lines,  which  repre- 
sent respectively  visible  and  invisible  edges.  The  correct 
representation  of  these  visible  and  invisible  edges  is  an  essen- 
tial part  of  the  projection  of  any  solid.  The  student's  facility 
in  determining  visibility  is,  to  a  considerable  extent,  a  measure 
of  his  understanding  of  the  problem. 

In  visualizing  any  object  or  objects,  an  /T-projection,  or  plan, 
is  always  viewed  from  above  ;  a  P~-projection,  or  elevation,  from 
hi  front  (see  §  8).  This  is  true,  whatever  be  the  relative  posi- 
tions of  these  projections  with  respect  to  the  ground  line,  or  to 
each  other. 

In  a  drafting  office,  the  relative  position  of  plan  and  eleva- 
tion is  usually  prescribed  by  the  custom  of  the  office,  so  that 
it  is  known  by  the  position  on  the  sheet  which  view  is  to 
be  read  as  an  //"-projection,  and  which  as  a  P"-projection. 
In  studying  the  theory,  however,  this  is  not  the  case  ;  objects 
may  be  placed  in  any  position  in  any  quadrant,  and  some 
indication  as  to  which  is  the  l/-p  rejection  and  which  the  V* 
projection  must  always  be  given.  For 
the  present,  we  shall  do  this  by  the  nota- 
tion, using  the  index  letters  A  and  •  (§  12). 

25.  Illustrative  Examples.  The  stu- 
dent should  visualize  the  solids  projected 
in  Figs.  26-32,  aided  by  the  following 
hints.  Notice  carefully  the  visibility  in 
each  case,  as  shown  by  the  full  and  dotted 
lines  (§  24). 

Figure  26.  First  quadrant.  A  square 
right  prism.  The  base  is  in  H,  and  the 
lateral  edges  are  vertical.  The  //-projec- 
tion shows  the  true  size  of  the  base,  the 
distance  of  each  lateral  edge  from  V,  and 
the  angle  between  each  lateral  face  and  V. 
shows  the  altitude  of  the  prism. 


Fig.  26. 


The  F-projection 


16  DESCRIPTIVE  GEOMETRY  [II,  §  25 

Figure  27.  Third  quadrant.  A  rectangular  right  prism.  The 
base  is  parallel  to  H,  and  the  lateral  edges  are  vertical.  The 
//-projection  shows  the  true  size  of  the  base,  and  its  position 
(angles,  distances)  with  respect  to  F  The  F-projection  shows 
the  altitude  of  the  prism  and  its  distance  from  H. 

Figure  28.  Fourth  quadrant.  A  triangular  right  prism. 
The  base  is  parallel  to  V,  and  the  lateral  edges  are  per- 
pendicular to  V.  The  F-projection  shows  the  true  size  and 
shape  of  the  base,  and  its  position  with  respect  to  H.  Note 
that  the  base  is  not  a  regular  (equilateral)  triangle.  The 
//-projection  shows  the  length  of  the  prism  and  its  distance 
from  V. 

Figure  29.  Second  quadrant.  A  square  right  pyramid.  The 
base  is  parallel  to  H.  The  //-projection  shows  its  size  and 
position.  The  F-projection  shows  the  distance  of  the  base  from 
H,  and  the  altitude  of  the  pyramid. 

Figure  30.  Third  quadrant.  A  pentagonal  right  pyramid. 
The  base  is  parallel  to  H.  The  apex  is  below  the  base,  so  that 
the  pyramid  is  inverted.  The  base  is  a  regular  pentagon.  One 
edge  of  the  base  is  perpendicular  to  F,  so  that  one  of  the  lateral 
faces  projects  on  Fas  a  straight  line. 

Figure  31.  First  quadrant.  A  frustum  of  a  regular  hexag- 
onal pyramid.  The  lower  base  of  the  frustum  is  in  H,  the 
upper  base  is  parallel  to  H.  In  order  to  aid  in  the  visualization 
of  the  frustum,  the  complete  pyramid  is  shown  in  light  dotted 
lines. 

Figure  32.  Quadrant  indeterminate.  An  irregular  triangular 
pyramid.  This  solid  may  be  constructed  by  assuming  any  four 
points  in  space,  and  then  connecting  each  point  with  the  other 
three,  thus  forming  a  solid  bounded  by  four  triangular  faces. 
In  visualizing  this  pyramid,  any  one  of  the  four  points  may  be 
taken  as  the  vertex,  and  the  other  three  points  as  the  corners 
of  the  base. 

In  Fig.  32,  no  ground  line  is  shown;  hence  the  distances 
of  the  object  from  H  and  F  cannot  be  told.  Nevertheless,  the 
projections  represent  a  pyramid  of  definite  size  and  shape 
(compare  §  14,  Figs.  15  and  16). 


II,  §  25]  POINTS  AND  SIMPLE  SOLIDS 


17 


Fig.  27. 


Fig.  28. 


Fig.  29. 


Fig.  30. 


Fig.  31. 


Fig.  32. 


18 


DESCRIPTIVE  GEOMETRY 


[II,  §  26 


26.  Curved  Solids.  In  projecting  solids  bounded  partly  or 
wholly  by  curved  surfaces,  the  projection  cannot  always  be 
made  wholly  of  edges  of  the  solid,  but  may  consist  partly 
or  wholly  of  the  apparent  outline  or  contour  of  the  object. 

A  few  typical  curved  solids  are  represented  in  Figs.  33-38. 


Fig.  33. 


Fig.  34. 


Fig.  35. 


Fig.  36. 


No  ground  line  is  shown,  but  the  objects  are  all  supposed  to  be 
in  the  first  quadrant : 

A — A  circular  right  cylinder. 

B  —  A  circular  right  cone. 


cm 


C — A  sphere. 

D  —  A  hollow  hemi- 
sphere, or  hemispherical 
bowl. 

E  —  A  torus.  This  is  the 
solid  generated  by  revolv- 
ing a  circle  about  an  axis 
situated  in  its  own  plane, 
but  outside  the  circle. 

F — A  spool,  consisting 
of  a  cylindrical  body,  coni- 
cal ends,  and  with  a  cy- 
lindrical hole  lengthwise 
through  it. 


CHAPTER   III 
REPRESENTATION   OF   THE   STRAIGHT  LINE  —  TRACES 

27.  The  Straight  Line.  A  line  is  the  path  of  a  moving  point, 
and  is  not  necessarily  straight.  Yet  in  ordinary  use,  the  term 
line,  by  itself,  and  without  anything  to  imply  the  contrary, 
always  means  a  straight  line. 

28.  Projections  of  the  Straight  Line  (Fig.  39).  Let  cd  be  any 
straight  line,  and  Q  any  plane  of  projection.     Then  as  a  point 


Fig.  [59. 

moves  along  the  line  from  c  to  d,  its  projection  will  move  along 
the  plane  from  cfl  to  d".    The  following  propositions  are  evident : 

1.  TJie  projection  of  a  straight  line  is  a  straight  line. 

2.  Tlie  projection  of  any  point  in  the  line  lies  in  the  projection 
of  the  line. 

Since  any  point  in  space  is  definitely  determined  when  its 
projections  on  //and  Tare  known,  it  follows  that,  in  general, 
any  two  straight  lines  assumed  at  random,  one  in  //  and  one  in 
V,  are  the  projections  of  one  and  only  one  straight  line  in 
space.     Certain  exceptions  will  be  noted  in  §  34. 

29.  Notation  of  the  Straight  Line.  A  straight  line  of  definite 
length  will  be  given  by  the  two  points  at  its  extremities,  as  the 
line  cd  (chdh,  cdv).  A  straight  line  (if  indefinite  length  will  be 
denoted  by  a  single  capital  letter.  This  letter  alone  denotes 
the  line  in  space,  and  the  letter  with  a  suitable  index  indicates 
a  projection.     Thus  we  shall  say  tin-  line  A,  or  the  line  Ii(Bh,  B"). 

19 


20  DESCRIPTIVE  GEOMETRY  [III,  §  30 

30.  Visualizing  the  Straight  Line.  The  following  example 
illustrates  a  general  method  of  visualization.  A  single  projec- 
tion, as  we  have  seen,  is  not  sufficient  to  locate  a  point  or  line 
in  space.  By  a  constant  comparison  of  one  projection  with  the 
other,  however,  a  sufficient  number  of  facts  are  brought  out  to 
complete  the  mental  picture  of  the  conditions  in  space. 

Consider  the  line  A,  Fig.  40.  In  visualizing  or  "  seeing  "  the 
line  in  space,  either  the  //"-projection  or  the  F-projection  may 
be  used  first  as  a  basis.  As  the  natural  position  of  the  drawing 
surface  is  horizontal,  let  the  //-projec- 
tion  be  chosen  first.  Then  in  viewing 
the  line  A  from  above,  the  line  in  space 
is  either  directly  over  or  directly  under 
Ah,  or  else  partly  above  and  partly  below. 
Visualizing  in  this  direction  alone,  how- 
ever,  can  give  no  idea  whether  the  line 
is  parallel  or  oblique  to  the  //-plane.  By  now  looking  toward 
V,  and  seeing  the  line  in  space  projecting  as  Av,  the  fact  is 
revealed  that  the  left-hand  end  of  the  line  is  the  higher,  and 
that,  reading  from  this  end,  the  direction  or  slope  of  the  line 
in  space  is  downward  to  the  right.  Returning  to  the  if-projec- 
tion,  and  reading  along  the  line  in  the  same  direction,  namely, 
from  left  to  right,  the  line  is  seen  to  slope  away  from  the 
observer,  that  is,  backward.  Finally,  by  comparison  of  the 
two  projections,  we  find  that  the  left-hand  end  of  the  line  is  in 
the  first  quadrant,  the  right-hand  end  in  the  third,  the  central 
portion  in  the  second  quadrant  (§§  16,  19). 

31.  The  Slope  of  a  Line.  In  determining  the  slope  of  a  line, 
the  directions  up  and  down,  right  and  left,  both  of  which  are 
shown  by  the  ^projection,  rarely  trouble  the  student.  The 
directions  forward  and  backward,  however,  which  are  alwajrs 
shown  by  the  //"-projection,  often  cause  confusion.  This  diffi- 
culty should  disappear  by  considering  the  method  of  projecting 
a  solid  object ;  for  in  projecting  any  actual  object,  the  side 
nearest  the  observer  is  always  considered  to  be  the  front  of  the 
object  (see  §§  20,  24).     A  point  which  is  farther  away  from 


HI,  §  32] 


STRAIGHT  LINES 


21 


Fig.  41. 


the  observer  is  behind  one  which  is  nearer ;  so  that  the  direc- 
tion from  front  to  back  is  away  from  the  observer.  This  is 
shown  in  Fig.  41  by  the  arrow 
A,  which  points  away  from  the 
observer,  and  therefore  points 
backward.  The  arrow  B  in  this 
figure  points  toward  the  ob- 
server, and  is  pointing  forward. 
Points  in  the  first  and  fourth 
quadrants  are  always  in  front  of 
points  in  the  second  and  third 
quadrants ;  but  points  on  the 
same  side  of  the  F"-plane  must 
be  read  by  the  relative  positions  of  their  .^-projections,  the 
lower  side  of  an  iZ-projection  being  always  the  front. 

The  slope  of  the  line  A,  Fig.  40,  would 
be  described  as  downward,  backward,  to 
the  right.  The  slope  may  also  be  given  as 
upward,  forward,  to  the  left,  since  it  is  im- 
material in  which  direction  the  line  is  read. 

32.  Additional  Examples.  Line  B,  Fig. 
42.  Eead  in  the  manner  already  described, 
it  will  be  found  that  this  line  is  oblique  to  H  and  V,  passes 
through  the  fourth,  third,  and  second  quadrants,  and  slopes 
upward,  backward,  to  the  right  (or 

downward,  forward,  to  the  left).  

Line  A,  Fig.  43.  The  F-projection 
is  parallel  to  the  ground  line,  so  that 
every  point  of  the  line  is  the  same  dis- 
tance from  II.  The  line  A  is  parallel 
to  II,  oblique  to  V,  passes  through 
the  second  and  first  quadrants  only,  Flr;-  43- 

and  slopes  forward  to  the  right  (backward  to  the  left). 

Note.  Lines  parallel  to  //are  known  as  ^-parallels  ;  lines  parallel  to 
V  are  known  as  F-parallels.  A  line  parallel  to  //  may  also  be  called  a 
horizontal  line  :  but  the  corresponding  term,  vertical  line,  is  limited  to  a 
line  perpendicular  to  H.     h-  and  F-parallels  are  very  important  lines. 


Fig.  42. 


22 


DESCRIPTIVE  GEOMETRY 


[HI,  §  33 


Fig.  45. 


33.  Special  Positions  of  the  Line.     The  student  should  satisfy 
himself  of  the  truth  of  the  following  propositions  by  direct 

visualization. 

1.  If  a  line  A  intersects 
the  ground  line,  Ah  and  Av 
intersect  on  GL,  and  in  the 
same  point  as  A  (Fig.  44). 

2.  Conversely,  if  Ah  and 
FlG-  4i-                               jiy  intersect  the  ground  line 

in  the  same  point,  the  line  A  intersects  the  ground  line.     Through 
how  many  quadrants  does 
such  a  line  pass  ? 

3.  If  a  line  A  is  parallel 
to  H,  Av  is  parallel  to  the 
ground  line;  and  conversely, 
if  Av  isimrallel  to  the  ground 
line,  A  is  parallel  to  H  (Fig. 
43). 

4.  If  A  is  parallel  to  V,  Ah  is  parallel  to  the  ground  line;  and 
conversely,  if  Ah  is  parallel  to  the  ground 
line,  A  is  parallel  to  P"(Fig.  45). 

5.  If  A  is  perpendicular  to  H,  Av  is 
perpendicular  to  the  ground  line,  and  Ah  is 
a  point;  conversely,  if  Ah  is  a  point,  A  is 
perpendicular  to  H.     (See  line  A,  Fig.  46.) 

6.  If  A  is  perpendicular  to  V,  Ah  is 
perpendicidar  to  the  ground  line,  and  A" 
is  a  point;  conversely,  if  A"  is  a  point,  A  is 
perpendicular  to  V.     (See  line  B,  Fig.  46.) 

7.  If  A  lies  in  H,  Ah  coincides  with  A,  while 
Av  coincides  icith  the  ground  line  (Fig.  47). 
The  statements  of  the  analogous  and  converse 
propositions  are  left  to  the  student. 

34.  Exceptional  Cases.      While   in   general 
(§  28)  the  two  projections  of  a  line  may  be 
assumed   in  any  direction  at  random,  certain  exceptions  are 
illustrated  by  the  following  examples  : 


■  ■Bv 


+  An 
Fig. 


46. 


Fig. 


Ill,  §  34] 


STRAIGHT  LINES 


23 


(a)  (Fig.  48).  The  projection  Ah  is  perpendicular  to  the 
ground  line,  while  the  projection  Av  is  not.  Assume  on  Ah 
any  point  ch,  and  find  cv  by  projecting  to  Av.  The  projector 
chcv  coincides  with  Ah,  hence  cv  is  always  the  same  point  on  A", 
no  matter  where  ch  is  chosen.  Now  let  d°  be  any  point  on  Av 
except  C.     The  projector  from  dv  is  parallel  to  Ah,  and  there 


■  ■cT 


Fig.  48. 


•  ■rh 


Fig.  49. 


o 


TcT 


leh 


Fig.  50. 

is  no  point  dh  to  correspond  with  d°.     Hence  there  is  no  line  in 
space  to  correspond  with  the  given  projections. 

An  analogous  case  arises  if  Av  is  taken  perpendicular  to  the 
ground  line,  while  Ah  is  not. 

(b)  (Fig.  49).  The  projections,  Bh  and  Bv,  are  both  perpen- 
dicular to  the  ground  line,  but  at  different  points.  Let  ch  be 
any  point  on  Bh ;  the  projector  from  ch  is  parallel  to  Bv,  and 
there  is  no  point  on  B"  to  correspond  with  c\  Similarly,  if  dv 
is  any  point  on  Bv,  the  projector  from  d"  is  parallel  to  Bh,  and 
there  is  no  point  dh.  Therefore  there  is  in  space  no  line  B 
corresponding  to  these  two  projections. 

(c)  (Fig.  50).  The  projections  O1  and  Ovare  both  perpendicu- 
lar to  the  ground  line,  and  at  the  same  point.  Let  a/"  be  any 
point  on  Ch.  The  projector  from  ah  coincides  with  O,  and  the 
particular  point,  a",  on  C",  which  corresponds  with  ah  cannot 
be  determined.  The  line  C  is  therefore  indeterminate.  Bui 
if  the  line  be  projected  by  means  of  two  of  its  points  (§  29),  it 
at  once  becomes  determined.  Thus  the  line  de,  Fig.  50,  is  a 
definite  line,  lying  in  a  plane  perpendicular  to  the  ground  line. 


24 


DESCRIPTIVE  GEOMETRY 


[III,  §  35 


35.  The  Profile  Line.  A  plane  which  is  perpendicular  to 
the  ground  line  is  known  as  a  profile  plane,  and  any  line  lying 
in  a  profile  plane  is  termed  a  profile  line.  We  have  just  seen 
that  such  a  line  cannot  be  projected  in  the  same  way  as  the 
general  straight  line.  Hence  problems  in  which  such  lines 
occur  will  usually  —  but  not  always  —  call  for  particular  solu- 
tions. The  simplest  method  of  dealing  with  a  profile  line  is 
usually  by  means  of  an  additional  plane  or  projection.  The 
solution  of  cases  involving  such  lines 
will  be  deferred,  in  general, until  Chap- 
ter VI,  where  this  topic  is  considered. 
36.  A  Point  Lying  in  a  Line  (Fig. 
51).  It  follows  at  once  from  the 
second  proposition  of  §  28,  that  if  a 
point  lies  in  a  line,  the  projections 
of  the  point  lie  in  the  projections 
FlG-  51-  of  the  line.     This  condition  is  suffi- 

cient if  the  line  does  not  lie  in  a  profile  plane.     (See  §  35.) 

37.  Traces  of  a  Line.  Of  all  the  points 
in  a  straight  line,  the  two  in  which  it 
pierces  the  planes  of  projection  are  con- 
sidered the  most  important.  These 
points  are  called  the  traces  of  the  line. 
The  horizontal  trace  is  the  point  in  which 
the  line  pierces  H,  and  will  be  designated 
as  the  point  s ;  the  vertical  trace,  in  which 
the  line  pierces  V,  will  be  called  t. 

Problem  1.     To  find  the  traces  of  a  straight  line. 
Analysis.     The  solution  of  this  problem  depends  on  direct 
visualization. 
General  Case.     Line  not  lying  in  a  profile  plane. 
Construction  (Figs.  52,  53).      The  horizontal  trace.     Looking 
towards  V,  H  is  seen  edgewise  as  GL,  the  line  A  appears  as 
A",  hence  the  line  A  will  pierce  H  at  the  point  seen  as  s", 
where  Av  crosses  GL.     "While  this  projection  conveys  no  idea 
of  the  distance  of  the  point  in  //  from  V,  it  does  single  out,  to 


Fig.  52. 


Ill,  §  38] 


STRAIGHT    LINES 


25 


the  exclusion  of  every  other  point,  the  H  piercing  point  of  the 
line.  The  actual  position  of  the  point  in  H  is  found  at  sh  by 
the  projector  sl'sh. 

The  vertical  trace.  Looking  down 
toward  H,  V  is  seen  edgeAvise  as  GL,  A 
is  seen  as  Ah,  hence  A  is  seen  to  pass 
through  V  at  the  point  th.  The  actual 
location,  t",  in  V,  is  found  on  Av  by 
means  of  the  projector  thtv. 

Special  Case.  A  profile  line  (no  figure).  The  general  solution 
evidently  fails  ;  the  solution  will  be  given  in  Chapter  VI. 

Problem  2.  Given  the  two  traces  of  a  straight  line,  to  find  its 
projections. 

Analysis.  Each  trace  is  a  point  lying  in  a  coordinate  plane ; 
and  for  such  a  point,  one  projection  is  identical  with  the  point, 
the  other  projection  is  in  the  ground  line. 

Construction  (Figs.  52  and  53).  The  traces  sh  and  V  are  given ; 
sv  and  th  are  found  by  projection  on  GL ;  s* 
and  th  determine  Ah,  sv  and  V  determine  Av. 

If  sh  and  tv  lie  in  the  same  projector  (Fig. 
54),  the  required  line  is  a  profile  line.  Note 
that  this  is  a  definite  line,  since  two  points, 
s  and  t,  are  known. 

38.  Exceptional  Cases.  The  solution  to 
Problem  2  shows  that,  in  general,  a  straight 
line  is  uniquely  determined  by  its  two  traces. 
The  only  exception  is  the  case  in  which  s  and  t 
fall  together  as  a  single  point  on  the  ground 
line.  The  required  line  can  then  be  any  line 
which  passes  through  this  point;  in  order  to 
determine  the  line,  some  other  condition  must  be  given. 
The  following  questions  on  traces  are  left  to  the  student : 

1.  How  must  a  line  be  placed  so  as  to  have  but  one  trace  ? 

2.  Hew  must  a  line  be  placed  so  as  to  have  no  truer  ? 

3.  How  must  a  line  be  placed  so  that  its  horizontal  (or  ver- 
tical) trace  is  indeterminate  ? 


-■tv 


X 


Fig.  54. 


CHAPTER   IV 


SIMPLE    SHADOWS 


39.  Shadows.  The  shadow  of  a  point  on  a  plane  is  the  point 
in  which  a  ray  of  light  passing  through  the  given  point  is  in- 
tercepted by  the  plane. 

The  conventional  ray  of  light  is  a  line  sloping  downward, 
backward,  to  the  right,  with  both  projections  inclined  at  45° 
with  the  ground  line.  Such  a  ray  is  shown  by  the  line  L, 
Fig.  55. 

40.  Shadow  of  a  Point.  The  shadow  on  V  of  the  point  a, 
Fig.  55,  is  the  point,  aa  in  which  the  ray  of  light,  L,  passing 

through  a,  intersects  V.  The  shadow 
on  H  is  the  point,  as,  where  L  pierces  H. 
It  will  be  noted  that  the  points  at 
and  as  are  the  traces  (§  37)  of  the  line 
L  on  V  and  H  respectively.  Hence 
the  shadow  of  a  point  on  the  coordinate 
planes  is  found  by  drawing  through 
the  point  a  line  representing  the  ray 
of  light,  and  then  finding  the  traces 
of  this  line. 


Fig.  55. 


41.  Opaque  and  Transparent  Planes. 
In  order  that  an  actual  shadow  may 
be  cast,  it  is  necessary  that  the  plane 
receiving  the  shadow  be  opaque.  Thus,  in  Fig.  55,  if  V  be  con- 
sidered opaque,  at  is  the  actual  shadow  of  the  point  a ;  but  if 
V  be  considered  transparent  and  H  opaque,  as  will  be  the 
actual  shadow.  In  work  with  shadows,  it  is  customary  to 
place  the  object  in  the  first  quadrant,  and  to  consider  both 
//and  'Fas  opaque. 

Since  the  two  projections  of  the  ray  of  light  make   equal 
angles  with  the  ground  line,  it  follows  that  the  actual  shadow 

20 


IV,  §  43] 


SIMPLE  SHADOWS 


27 


of  a  point  must  always  fall  on  the  nearer  coordinate  plane.  In 
the  case  of  a  solid  object,  the  actual  shadow  may  fall  wholly  on 
H,  wholly  on  V,  or  partly  on  both  planes,  according  to  the  size 
and  position  of  the  object. 

42.  Shadow  of  a  Line.  The  shadow  of  a  line  will  consist  of  the 
shadows  of  the  points  which  compose  the  line.  If  the  line  is 
straight,  only  the  shadows  of  the  two  ends  of  the  line  need  be 
cast.  In  Fig.  56,  the  actual  shadows  of  both  points  a  and  b  fall 
on  V ;  hence  the  shadow  of  the  line  ab  is  the  line  atbn  lying 


wholly  in  V.  In  Fig.  57,  the  actual  shadow  falls  partly  on  // 
and  partly  on  V.  To  find  this  shadow,  Ave  may  find  the  com- 
plete shadow  on  each  plane,  regarding  the  other  plane  as  trans- 
parent, and  then  take  the  actual  portion  of  each  shadow.  But 
the  two  shadows,  asbs  and  atbt,  one  in  //  and  one  in  V,  must  in- 
tersect in  a  point,  e,  in  GL,  since  this  line  is  the  intersection  of 
the  planes  //  and  V.  Hence  we  may  find  one  complete  shadow, 
as  a  b,  note  the  point  e  in  which  ajb,  crosses  GL,  and  draw  from 
e  to  the  actual  shadow  at. 

43.  Shadows  of  Surfaces.  Shadows  of  surfaces  and  solids  are 
found  by  extending  the  methods  just  given  for  the  point 
and  straight  line.  The  only  other  case  which  will  be  taken 
up  at  this  time  is  that  of  a  convex  solid  bounded  by  plane 
faces. 


28 


DESCRIPTIVE  GEOMETRY 


[IV,  §  44 


44.  Shadow  of  a  Convex  Solid.  A  frustum  of  an  irregular 
triangular  pyramid  (Eig.  58)  has  six  corners  and  nine  rectilinear 
edges.  Not  all  of  them  cast  shadows  which  lie  on  the  boundary 
of  the  complete  shadow.  Find  the  shadows  of  the  six  corners  1, 
2  ...  6  on  one  of  the  planes,  as  II,  and  draw  the  largest  convex 


%—* 


Fig.  58 


polygon  possible,  with  three  or  more  of  these  shadow  points  as 
vertices.  In  Fig.  58,  the  shadow  is  the  polygon  14245s6,441j> 
the  point  3,  being  omitted  because  it  falls  inside  the  polygon. 
The  part  lse/4s  on  H  in  front  of  GL  is  an  actual  shadow ;  the 
remaining  part,  e2,sbJosf,  shows  the  portion  of  the  shadow 
intercepted  by  V.  Find  the  actual  shadows  on  Voi  the  points 
2,  5,  and  G  ;  then  e2l5filf  is  the  part  which  falls  on  V,  and 
completes  the  actual  shadow. 


CHAPTER   V 
REPRESENTATION   OF   THE   PLANE 

45.  Representation  of  the  Plane.  It  has  been  shown  (§  28) 
that  a  straight  line  of  indefinite  extent  can  be  represented  by- 
two  orthographic  projections  on  H  and  on  V. 

This  method,  however,  cannot  be  extended  to  the  representa- 
tion of  a  plane  of  indefinite  extent,  since  the  projections  of 
all  the  points  in  the  plane  would,  in  general,  cover  the  whole 
plane  of  projection. 

It  has  been  shown,  however  (§  38),  that  a  line  of  indefinite 
extent  may  also  be  determined  by  its  points  of  intersection 
with  the  planes  of  projection. 

This  method  is  capable  of  extension  to  the  representation  of 
a  plane. 

46.  Traces  of  a  Plane.  A  plane  of  indefinite  extent  will,  in 
general,  cut  H  and  V  in  two  straight  lines,  which  meet  in  the 
point  in  which  the  given  plane  cuts  the  ground  line. 

These  lines  of  intersection  of  the  plane  with  It  and  V  are 
called,  respectively,  the  horizontal  and  vertical  traces  of  the  given 
plane. 

Conversely,  two  arbitrary  straight  lines,  one  in  H  and  one  in 
V,  which  intersect  the  ground  line  at  the  same  point,  determine 
a  plane  of  which  the  arbitrary  lines  are  the  horizontal  and  ver- 
tical traces. 

Thus  a  definite  method  of  representation  of  the  plane  has 
been  found. 

This  method  of  representation  fails  only  in  the  case  in  which 
the  plane  contains  the  ground  line,  when  the  two  traces  will 
coincide  with  each  other  and  with  the  ground  line.  In  this 
case  the  plane  will  not  be  determined  until  some  additional 
condition  is  imposed. 

Compare  this  with  the  analogous  case  of  a  line  having  coin- 
cident traces,  §  38, 

29 


30 


DESCRIPTIVE  GEOMETRY 


[V,  §  47 


47.  Notation  of  the  Plane.  A  plane  in  space  will  be  denoted 
by  a  single  capital  letter,  taken  usually  from  the  middle  or 
end  of  the  alphabet. 

The  horizontal  trace  of  a  plane  Q,  being  the  line  of  intersec- 
tion of  H  and  Q,  will  be  called  HQ. 

The  vertical  trace,  or  intersection  of  V  and  Q,  will  be  called 
VQ. 

The  form  HQ  is  used  in  preference  to  the  form  QH,  since  the 
latter,  in  speech,  might  be  confused  with  Qh,  the  //"-projection 
of  a  line  Q,  which  is  by  no  means  the  same  thing  as  the  //-trace 
of  a  plane  Q. 

48.  Visualization  of  the  Plane.  The  visualization  of  a  plane 
from  its  traces  is  an  entirely  different  process  from  reading 
the  projections  of  a  line  or  of  a  point.  The  traces  of  a  plane 
oblique  to  H  and  V  are  not  projections  of  the  plane  in  the 
ordinary  sense. 

Let  HQ  and  VQ,  Fig.  59,  be  the  horizontal  and  vertical  traces 
of  a  given  plane  Q. 


V 

/ 

/  /  \/ 

V      V 

/%> 

\ 

^           /                      / 

N      /                      / 
\/                      / 

Fig.  59. 

Fig.  60. 

I 


Since  the  plane  is  indefinite  in  extent  and  cuts  the  ground 
line,  it  must  pass  into  all  four  quadrants. 

Since  the  two  traces  are  oblique  to  GL,  it  follows  that  the 
plane  determined  by  them  is  oblique  to  both  H  and  V. 

49.  The  Slope  of  a  Plane.  The  slope  of  an  oblique  plane  is  de- 
scribed in  the  same  terms  as  the  slope  of  a  line,  and  may  perhaps 
be  best  understood  by  reference  to  the  planes  of  Figs.  59—63. 


V,  §  49]       REPRESENTATION  OF  THE  PLANE 


31 


Plane  Q  in  Fig.  59,  shown  pictorially  in  Fig.  60,  would  be 
said  to  slope  downward,  forward,  and  to  the  left  (or  upward, 
backward,  to  the  right). 

In  Fig.  61  the  slope  is  downward,  backward,  to  the  left  (ab- 
breviated to  D.B.L.). 


Fig.  61. 


In  Fig.  62,  the  slope  is  downward,  forward,  to  the  right 
(D.F.R.). 


r 

/// 

p. 

\ 

t» 

V 

V 

f 

s 

Fig.  (J2. 


In  Fig.  63,  plane  Q,  perpendicular  to  H,  would  be  said  to 
slope  simply  backward  to  the  right  (B.R.). 


VR 


J. 


LJ" 


^ 


A, 


VR 


& 


Fig.  63. 


It  is  usually  easier  to  visualize  the  slope  of  a  plane  by  con- 
sidering, at  first,  one  quadrant  only  (the  first  or  third).  The  plane 
can  then  be  visualized  as  extending  into  the  other  quadrants. 


32 


DESCRIPTIVE   GEOMETRY 


[V,  §50 


50.  Special  Positions  of  the  Plane.  The  student  should  satisfy 
himself  of  the  truth  of  the  following  propositions  by  direct 
visualization. 

1.  If  a  plane,  Q,  is  parallel  to  H,  there  is  no  H-trace  (i.e.  HQ 
does  not  exist),  while  VQ  is  parallel  to  the  ground  line. 

2.  Conversely,  if  Q  has  no  H-trace,  VQ  must  be  parallel  to  the 
ground  line,  and  Q  is  parallel  to  H 

3.  Similarly,  if  Q  is  parallel  to  V,  VQ  does  not  exist,  while 
HQ  is  parallel  to  the  ground  line;  and  conversely. 

4.  If  Q  is  parallel  to  the  ground  line,  HQ  and  VQ  are  also 
parallel  to  the  ground  line  ;  and  conversely. 

5.  If  Q  is  perpendicular  to  H,  VQ  is  perpendicular  to  the 
ground  line  ;  if  Q  is  perpendicular  to  V,  HQ  is  perpendicular 
to  the  ground  line  ;  and  conversely  in  each  case. 

6.  If  Q  is  perpendicular  to  both  H  and  V,  that  is,  perpen- 
dicular to  the  ground  line,  HQ  and  VQ  fall  together  in  the  same 
perpendicular  to  GL. 

51.  Edge  View  of  a  Plane.  The  particular  position  of  a  plane 
in  which  it  becomes  perpendicular  to  one  of  the  coordinate 


VR 


& 


Fig.  63  (repeated). 


planes  is  an  important  one  to  visualize.  If  Q  is  perpendicular 
to  H,  then  the  //-projection  of  every  point  in  Q  must  fall  in 
HQ,  which  thus  becomes  an  actual  projection  of  the  plane.  The 
trace  HQ  is,  in  fact,  an  edge  view  of  the  plane  Q  (Fig.  63).  It 
should  be  noted  that  a  plane  parallel  to  one  of  the  coordinate 
planes  is  necessarily  perpendicular  to  the  other,  and  that  the 
single  trace  of  such  a  plane  is  its  edge  view  (plane  R,  Fig.  63). 


CHAPTER  VI 

THE   PROFILE   PLANE   OF  PROJECTION 

52.  The  Profile  Plane.  The  profile  plane,  when  used  as  a 
plane  of  projection,  will  be  designated  by  the  letter  P.  This 
plane  has  already  been  defined  as  a  plane  perpendicular  to  the 
ground  line  (§  35)",  that  is,  P  is  perpendicular  to  both  H  and 
V.  Hence  the  three  planes  H,  V,  and  P  form  a  system  of 
mutually   perpendicular   planes    (Fig.    64),   intersecting  in  a 


/ 

A 

V 

/ 
/     ■/ 
/     / 
/     / 

HV  / 

/ 

i 
1 

*    // 

/      / 
/      / 

A 

/ 

0 

U/~ 

ti-?-t 

x          /  4-i 

3 

i 
i 



i  / 

"7^ 

/ 

t 

7 

H  / 

1 L_ 

Fig.  04. 

common  point  o,  and  intersecting  each  other  by  pairs  in  three 
mutually  perpendicular  lines,  II V,  IIP,  and  VP,  all  passing 
through  o.    The  intersed  ion,  IIV,  of  //and  Fis  the  ground  line. 

53.    A  Profile  Projection.     Let   the  model  of    some  familiar 

object,  such  as  a  cabin,  be  placed  on  the  horizontal  plane  in 

tlic  first  quadrant,  l''i.^r.  <*>l.     Let  the  front  and  hack  be  parallel 

to  the  F-plane,  and  the  ends  parallel  to  P.     Let  a  view  be  taken 

D  ;j3 


34 


DESCRIPTIVE   GEOMETRY 


[VI,  §  53 


looking  in  the  direction  of  the  arrow,  that  is,  from  right  to  left. 
In  such  a  view,  the  whole  surface  of  P  will  be  seen,  while  the 


Fig.  m  (repeated). 
F-plane  will  be  seen  edgewise  as  a  vertical  line,  the  //-plane 
edgewise  as  a  horizontal  line,  and  only  the  right-hand  end  of 
the  cabin  will  appear. 

The  actual  end  view  is  shown  in  Fig.  65.     No  boundary  is 
shown  in  this  figure  for  the  P-plane,  since  it,  like  the  H-  and 


/\ 


P 
Fig.  65 


F-planes,  is  of  indefinite  extent.  The  quadrants  will  appear 
as  indicated  by  the  numbers,  and  must  always  so  appear  when 
P  is  viewed  from  the  right.     In  this  end  or  profile  view  of 


VI,  §  54] 


THE  PROFILE  PLANE 


35 


the  cabin,  both  heights  and  widths,  or,  in  other  words,  all  dis- 
tances perpendicular  to  //  or  to  V  are  shown  in  their  true 
size.  Hence  in  this  view  the  actual  distances  from  the  object 
to  H  and  to  V  are  necessarily  shown. 

54.    Relation  of   the  Profile   Projection  to   the   H-  and   7-Pro- 
jections.     Figure  66  shows  the  construction  of  the  end  view,  or 


ric 

Showing                            \J 
ht  hand  end 

AP 

< X * 

A' 

/rv"~: 

Front 

1 
1 

Back 

1 

£ 

•< 

H 

1 

\ 

\        \ 

i                    4- 

\ 
\ 
\ 
\ 

3 

Ah 

\ 
\ 
\ 

< 

V 



Fig.  66. 

profile  projection,  of  the  cabin  from  given  horizontal  and 
vertical  projections.  The  distance  x,  from  the  edge  view  of  V 
to  the  end  of  the  cabin,  may  be  chosen  at  will,  as  this  evidently 
has  no  effect  on  the  shape  or  size  of  the  end  view  of  the  cabin, 
or  its  distances  from  //  and  V. 

The  edge  views  of  //  and  V,  together  with  the  end  view  of 
the  cabin,  might  also  be  placed  at  the  right  of  the  plan  and 
elevation,  instead  of  the  left;  and  would  preferably  be  so 
placed  if  the  object  were  given  in  the  third  quadrant.     This 

would  locate  the  views  in  accordance  with  the  usual  custo t 

placing  the  view  of  the  right-hand  end  of  an  object  to  the 
right  of  its  front  view.  For  a  consideration  of  a  view  of  the 
left-hand  end,  see  §§  62  and  63. 


36 


DESCRIPTIVE  GEOMETRY 


[VI,  §  55 


55.  A  Second  Example.  Let  it  be  required  to  draw  the 
plan,  elevation,  and  side  elevation  (profile  projection)  of  a 
piano  packing  case  placed  in  the  third  quadrant,  with  its 
long  edges  parallel  to  both  H  and  V.  The  dimensions  of 
the  case  and  its  distances  from  H  and  V  are  supposed  to  be 


v        "x 

s      \ 
2 

\ 
\ 
\ 
\ 
\ 
\ 
\ 
\ 
\ 
i 

Plan 

H                 ' 

4 

3 

i 
i 

7 

.     t 

:  j 

View 

<—  m  — > 

of 

right 
hand 
end 

Elevation 

A" 

Fig.  67. 


known.  The  end  view  is  drawn  first  at  A"  (Fig.  67),  located 
as  shown  in  the  third  quadrant,  and  at  the  given  distances, 
k  from  H  and  m  from  V.  The  plan  and  elevation  may  then 
be  readily  constructed  as  shown.  It  is  important  to  see 
clearly  why  all  the  long  edges  appear  visible  both  in  plan  and 
elevation. 

56.  Profile  Projections  in  All  Four  Quadrants.  Figures  68-71 
show  a  regular  triangular  prism  placed  successively  in  the 
four  quadrants.  The  long  edges  of  the  prism  are  parallel  to 
both  H  and  V,  and  in  each  quadrant  the  upper  front  lateral 
face  makes  an  angle  of  15°  with  //.  In  each  position  the  right- 
hand  end  is  shown  in  the  profile  view.  The  student  should 
visualize  the  prism  for  each  of  the  four  positions,  noting  espe- 
cially the  visibility  of  the  edges  lettered  Bv  and  C\  lettered 
also  in  the  profile  view  as  the  points  B"  and  Cp. 


VI,  §  56]  THE  PROFILE   PLANE 


37 


2d  Quadrant, 


Fig.  68. 


Fig.  69. 


3d   Quadrant 


\    \ 
\    \ 


4-th    Quadrant 


"I r~r 


l       ,15' 


Fia.  70. 


Fig.  71. 


38 


DESCRIPTIVE   GEOMETRY 


[VI,  §  57 


Q. 
> 

bp 

bv 

a"7 

V 

1 
l 
i 

1 

i 

1 

2N 

i 

1 

•bh 

I 

H  : 

1 

1 

1 

S    i 

\ 

,4 

a. 

I 

3 

1    s 

1    s 

ah 

Fig.  72. 


57.  Profile  Projection  of  the  Straight  Line.  The  profile  pro- 
jection of  any  straight  line  may  be  obtained  from  its  hori- 
zontal and  vertical  projec- 
tions by  finding  the  profile 
projections  of  any  tAvo 
points  in  the  line  (Fig.  72). 

An  important  case  of  the 
profile  projection  of  a  line 
arises  when  the  line  is  par- 
allel to  P,  that  is,  when  the 
line  is  a  profile  line.  By 
means  of  its  profile  projec- 
tion we  can  solve  the  follow- 
ing problems  for  such  a  line,  which  cover  solutions  previously 
omitted  (§§  36,  37). 

58.  Problems  on  the  Profile  Line.  Given  the  horizontal  and 
vertical  projections  of  a  profile  line,  to  find 

(a)  its  true  length  ; 

(b)  its  traces  on  H  and  V ; 

(c)  the  projections  of  a  point  lying  in  the  line. 

(a)  The  true  lesgtii. 

Analysis.  Since  the  line  is  parallel  to  P,  its  profile  pro- 
jection will  be  equal  to  its  true  length. 

(b)  The  traces  ox  H  and  V. 

Analysis.  These  traces  will  appear  in  the  profile  projection 
as  the  points  in  which  the  line  cuts  the  respective  edge  views 
of  H  and  V. 

Construction  for  (a)  and  (b)  (Fig.  73).  Let  ab(ahbh,  avbv)  be 
the  given  profile  line.  If  the  profile  plane  of  projection  is  not 
given,  it  may  be  assumed  anywhere,  perpendicular  to  the 
ground  line.  Find  the  profile  projections  of  the  points  a  and 
b.  Then  apbp  is  the  profile  projection  of  the  line,  and  is  equal 
to  the  true  length  of  ab.  Produce  apb"  indefinitely.  Then  the 
line  is  seen  to  pierce  i/at  sp,  and  Fat  t".  Find  the  horizontal 
and  vertical  projections  of  the  points  s  and  t ;  the  point 
s  (sh,  sv)  is  the  H-tra.ce,  and  the  point  t  (th,  tv)  is  the  P~-trace. 


VI,  §  58] 


THE  PROFILE   PLANE 


39 


A  second  example  is  shown  in  Fig.  74.  This  is  the  more 
common  construction,  the  profile  plane  of  projection  being 
taken  directly  through  the  given  line. 

(c)   The  projections  of  a  point  on  the  line. 

Analysis.     If  the  profile  projection  of  the  line  be  found,  we 


Fig.  73. 


can  project  a  point  in  the  ^-projection  to  the  P-projection,  and 
from  this  to  the  F-projection ; 
or  vice  versa. 

Construction  (Fig.  75).  Let 
ab  (ahbh,  avb")  be  the  given 
line.  Find  apb".  Now  if  ch  is 
given,  project  from  c*  to  cp  in 
a"bp;  and  from  cp  to  C  in  <i'f>r. 
If  c"  is  given,  c*  may  be  found 
by  reversing  the  process. 

The  line  o.b  may  be  pro- 
duced indefinitely,  and  the  as- 
sumed point  may  be  beyond 
the  limits  of  a  and  b.  Thus,  if 
(I"  is  taken.  dh  is  found  as 
shown,  and  the  point  d  lies  on  the  line  produced  into  the 
third  quadrant.     Or,  dh  may  be  given,  and  d"  found. 


40 


DESCRIPTIVE   GEOMETRY 


[VI,  §  59 


59.  Profile  Trace  of  a  Line.  The  general  line,  oblique  to  H, 
V,  and  P,  will  pierce  all  these  planes.  Hence,  besides  the 
horizontal  trace  s,  and  the  vertical  trace  t,  it  will  have  a  profile 
trace,  u,  which  must  lie  in  the  profile  projection  of  the  line. 

Given  the  horizontal  and  vertical  projections  of  a  general  line, 
to  find  its  profile  projection  and  profile  trace. 

Analysis.  The  profile  projection  of  any  straight  line  can  be 
obtained  from  its  horizontal  and  vertical  projections  by  finding 
the  profile  projections  of  any  two  points  in  the  line  (§  57). 
The  profile  trace  can  be  found  by  noting  the  point  in  which 
the  given  line  pierces  the  edge  view  of  P. 

Construction  (Fig.  76).  Ah  and  Av  are  the  given  projections. 
Find  the  points  s  and  t,  the  H-  and  P"-traces  of  the  line  A 
(Prob.  1,  §  37).  Project  s  to  sp  in  GL,  and  t  to  tp  in  VP;  s» 
and  V  determine  the   required    profile    projection  Ap.     Since 


Fig.  76. 


HP  and  VP  are  both  edge  views  of  P,  we  have  uh  and  uv  as 
the  H-  and  F-projections  of  the  point  u  ;  from  uh  and  uv  rind 
up,  the  actual  profile  trace.     As  a  check,  up  must  lie  in  A". 

Since  any  two  of  its  projections  are  sufficient  to  determine 
the  line  A,  we  have  the  following  proposition : 

Corollary.  Given  Ap  and  either  Ah  or  A";  to  find  the 
remaining  projection  and  the  three  traces  of  the  line. 


VI,  §  60] 


THE   PROFILE   PLANE 


41 


Analysis.  The  desired  result  may  be  obtained  by  reversing 
the  preceding  construction. 

The  construction  is  left  to  the  student. 

60.  Profile  Trace  of  a  Plane.  A  plane  not  parallel  to  P  will 
intersect  it,  and  the  line  of  intersection  is  known  as  the  profile 
trace  of  the  plane. 


Fio.  77. 


A  general  plane  Q  is  shown  in  Fig.  77,  intersecting  H,  V, 
and  P,  in  the  traces  HQ,  VQ,  and  PQ.  It  is  evident  that  HQ 
and  VQ  intersect  on  HV  (the  intersection  of  H  and  V,  or 
ground  line),  II Q  and  PQ  intersect  on  IIP,  and  VQ  and  PQ 


Fig.  78. 


intersect  on  VP.     Whence,  if  HQ  and  VQ  are  given,  PQ  may 
be  found  as  shown  in  Fig.  78. 


42  DESCRIPTIVE  GEOMETRY  [VI,  §  60 

The  profile  trace  of  a  different  plane,  R,  is  found  in  Fig  79. 


Fig.  79. 


The  profile  trace  of  a  plane  parallel  to  the  ground  line  is 
found  in  the  same  way,  as  shown  in  Fig.  80.     This  is  an  ini- 


V 

/ 

vs 

*y 

CL 

> 

r      H 

\ 

n 

I 

"V 

HS 

Fig.  80. 


portant  case,  for  the  plane  S  is  perpendicular  to  P,  and  conse- 
quently the  profile  trace  is  an  edge  view  of  the  plane  S. 


Fig.  81. 
In  Fig.  81  the  given  plane  T  passes  through  the  ground  line? 


VI,  §  61] 


THE  PROFILE  PLANE 


43 


and  is  determined  by  the  coincident  traces  HT,  VT,  and  the 
point  c  lying  in  the  plane.  To  tind  the  profile  trace  PT,  find 
first  the  profile  projection  cp,  of  the  point  c.  Then,  as  in  the 
preceding  case,  PT  is  an  edge  view  of  T,  and  therefore  passes 
through  cp  and  o.  It  will  be  seen  that  the  plane  T  is  definitely 
determined  by  its  three  traces  HT,  VT,  and  PT. 

Instead  of  being  determined  by  passing  through  the  point  c, 
the  plane  T  might  also  have  been  determined  by  stating  the 
quadrants  through  which  it  passes,  and  the  angle  between  T 
and  either  H  or  V.  Then  the  profile  trace,  PT,  could  have 
been  located  accordingly. 

From  Figs.  78-81  it  is  evident  that  the  three  traces  of  a 
plane  are  so  related  that  if  any  two  are  given,  the  third  can  be 
found.  The  cases  in  which  the  given  traces  are  the  profile 
trace  and  either  the  horizontal  or  vertical  trace,  to  find  the 
remaining  trace,  are  left  to  the  student. 

61.  Planes  Parallel  to  H  or  V.  A  plane  parallel  to  II  or  V 
has  no  trace  on  one  of  these  planes,  and  the  student  sometimes 
has  difficulty  in  constructing  the  profile  trace.  Such  a  plane 
may  be  regarded  as  a  limiting  case  of  the  plane.  S,  Fig.  80 ; 
but  the  result  may  be  obtained  by  direct  visualization,  as  fol- 
lows. Since  the  plane  is  perpendicular  to  P,  the  profile  trace 
will  be  its  edge  view.     Then  the  profile  projection  shows  the 


PQ 


vq 


Fio.  82. 


Fig.  83. 


edge  views  of  both   //  and  V,  to  the  proper  our  of  which  the 
plane  will  show  parallel. 

A  plane  parallel  to  Ilia  shown  in  Fi;-,r.  82  ;  one  parallel  to  V 
in  Fig.  83. 


44 


DESCRIPTIVE   GEOMETRY 


[VI,  §  62 


62.  Left-side  Views.  Hitherto,  the  profile  projection  has 
been  obtained  in  just  one  way.  Hence  the  student  might 
suppose  that  the  method  of  projecting  on  the  profile  plane  is  as 
invariable  as  that  of  projecting  on  H  or  V.  In  practice,  how- 
ever, this  is  not  the  case.  The  method  thus  far  adopted,  when 
applied  to  an  actual  object,  always  gives  a  view  of  its  right 
side ;   whereas  a  view  of  the  left  side  may  be  decidedly  prefer- 


a+ 


->     ef+- 


+  a° 


Fig.  84. 

able.  For  this  purjiose,  P  is  viewed  from  its  left  side,  that  is, 
from  left  to  right,  and  then  projected  so  as  to  bring  this  side 
into  the  drawing  surface.  This  method  is  shown  for  a  point  in 
the  first  quadrant  in  Fig.  84.  That  this  method  is  directly 
opposite  to  that  previously  used  is  shown  also  by  the  line  ab, 
Fig.  85.     Since  the  H-  and  F-p rejections  of  a  and  b  show  that 


z 

1               / 

3 

4    t 

1       / 

i 
I 
i 

bh- 

— " 

i  / 

i 

i 

bv- 



'b' 

/ 

Fig.  85. 


these  points  are  in  the  first  and  fourth  quadrants  respectively, 

the  quadrants  must  appear  in  the  profile  projection  as  marked. 

In  Fig.  86  the  profile  trace  of  the  same  plane  Q  is  obtained, 

first,  by  looking  from  right  to  left,  and  then  by  looking  from 


VI,  §  64] 


THE  PROFILE  PLANE 


45 


left  to  right.     Note  that,  since  HQ  is  at  45°  with  the  gi-ound 
line,  PQ  coincides  with  VQ  in  the  tirst  figure,  and  that  the 


Fig.  8(>. 

second  figure  is  much  the  clearer,  especially  if  some  further 
construction  is  to  be  made  with  the  plane  Q. 

Hereafter  we  shall  view  P  from  either  side ;  although,  other 
things  being  equal,  the  direction  from  right  to  left  will  be  pre- 
ferred. If  the  quadrants  are  numbered  in  every  profile  projec- 
tion, there  should  be  no  confusion,  whichever  view  is  used. 

63.  Conventional  Placing  of  Left-side  Views.  When  the  pro- 
file projection  of  a  solid  object  is  made  so  as  to  show  its  left 
side,  it  is  customary  to  place  this  projection  at  the  right  of  the 
front  elevation  when  the  object  is  located  in  the  first  quadrant, 
and  at  the  left  of  the  front  elevation  when  the  object  is  located 
in  the  third  quadrant.  This  is  the  reverse  of  the  method  of 
placing  the  view  of  the  right  side,  as  shown  in  Figs.  68  and  70. 
(See  §  54.)  Actual  objects  are  rarely  placed  in  the  second  or 
fourth  quadrants,  so  that  there  is  no  fixed  custom  for  locating 
such  views. 

64.  Second  Method  of  Obtaining  Profile  Projections.  Profile 
projections  may  be  constructed  also  by  treating  the  profile 
plane  as  a  secondary  plane  of  projection,  by  the  methods  of  the 
next  chapter.  Such  a  construction  is  less  common,  however, 
than  that  given  above,  since  the  resulting  position  of  the  profile 
projection  is  not  so  generally  satisfactory. 


CHAPTER   VII 
SECONDARY    PLANES    OF    PROJECTION 

65.  Secondary  Planes  of  Projection.  In  practical  work,  views 
of  objects  are  often  wanted  in  other  directions  than  those  which 
are  obtained  by  the  use  of  the  horizontal  and  vertical  coordinate 
planes.  In  fact,  it  often  happens  that  an  actual  object  cannot 
be  adequately  represented  by  a  simple  plan  and  elevation.  In 
the  theory,  also,  an  additional  projection  in  a  suitably  chosen 
direction  may  give  readily  a  solution  otherwise  difficult  to  ob- 
tain. Such  views,  or  projections,  are  obtained  on  secondary 
planes  of  projection,  taken  perpendicular  to  either  H  or  V,  and 
making  any  angle  whatever  with  the  other  coordinate  plane. 

Secondary  planes  of  projection  are  by  preference  taken  per- 
pendicular to  H,  for  while  any  plane  perpendicular  to  H  is 
vertical,  a  plane  perpendicular  to  V  and  oblique  to  H  is  neither 
vertical  nor  horizontal,  and  therefore  does  not  conform  to  either 
of  the  natural  directions  (§  6). 

66.  The  Profile  Plane  as  a  Secondary  Plane.  A  very  impor- 
tant secondary  plane  of  projection  is  the  profile  plane,  which  is 
perpendicular  to  both  H  and  V.  This  plane  has  just  been 
discussed  in  detail  in  the  preceding  chapter.  Indeed,  some 
writers  consider  this  plane  as  a  third  coordinate  plane  of  equal 
rank  with  the  horizontal  and  vertical  coordinate  planes.  In 
the  present  chapter,  the  profile  plane  will  be  considered  as  a 
special  case  of  a  plane  perpendicular  to  H. 

67.  Secondary  Ground  Lines.  The  method  of  projecting  on 
a  secondary  vertical  plane  is  shown  in  Figs.  87  and  88.  The 
secondary  plane  Vx  intersects  H  in  a  secondary  ground  line, 
GXLX,  which  may  be  at  any  angle  with  the  original  ground 
line  GL.  Since  the  projectors  aav  and  a<V,  Fig.  87,  are  both 
parallel  to  H,  the  distances  ave  and  a^j  are  equal,  both  being 
equal  to  aah,  the  distance  of  the  point  from  H. 

46 


VII,  §  68]     SECONDARY  PLANES  OF  PROJECTION       47 


68.  Principles  of  Secondary  Projections.  In  general  for  any 
secondary  vertical  plane  of  projection  we  have  the  following 
propositions  : 


Fig.  87. 


Fig.  88. 


1.  Tlie  plane  may  be  viewed  from  either  side,  irrespective  of 
the  ])osition  of  the  object,  but  the  direction  of  sight  must  always 
be  at  right  angles  to  the  plane  (§  5),  and  the  projectors  must  be 
drawn  at  right  angles  to  GXLX. 

2.  Distances  above  II  mast  be  laid  off  on  the  farther  side  of 
GXLX,  and  distances  below  II  on  the  near  side. 

3.  All  vertical  distances,  either  above  or  below  H,  remain 
unchanged. 

4.  Hence,  if  we  have  given  the  plan  and  elevation  of  any  point, 
whether  it  be  an  isolated  point,  one  end  of  a  line,  or  a  corner 
of  a  solid,  the  secondary  V-projection  may  be  obtained  as  follows  : 

(a)  Decide  from  which  side  the  secondary  F"-plane  is  to  he 
viewed. 

(6)  Draw  a  projector  from  the  //-projection  of  the  point  per- 
pendicular to  the  secondary  ground  line. 

(c)  Take  the  distance  of  the  point  from  II,  and  mark  it  off 
on  the  farther  side  of  GXLX  if  the  point  is  ahove  II,  and  on  the 
near  side  if  the  point  is  below.  Thus,  in  Fig.  87,  Vx  is  viewed 
in  the  direction  of  the  arrow,  and  the  projection  ax"bxv  can  In- 
laid off  only  on  the  side  of  GXLX  as  shown.  <  inversely,  if  the 
secondary  projection  axvbxv  is  constructed  as  in  Fig.  88,  it  follows 
that  Vx  is  viewed  from  the  side  indicated  by  the  arrow,  Fig.  87. 


48 


DESCRIPTIVE   GEOMETRY 


[VII,  §  69 


69.    Additional   Examples.     Additional   examples   of   projec- 
tions on  a  secondary  F-plane  are  given  in  Figs.  89  and  90.     In 


Fig.  89. 


Fig.  89,  note  that  the  directions  of  the  points  a  and  &  from 
H,  as  shown  by  the  relations  of  a"  and  bv  to  OL,  must  be  pre- 
served in  the  secondary 
projection.  In  Fig.  90  a 
profile  line  is  shown  pro- 
jected on  a  profile  plane 
by  this  method.  (See 
§§  64,  66.) 

70.  Simplification  of 
Problems  by  Means  of 
Secondary  Projections.  In 
the  solution  of  a  problem, 
there  is  no  advantage  in 
introducing  a  secondary 
plane  of  projection  unless 
the  new  projection  is  in 
some  way  simpler  than 
the  original  projections. 
A  point  always  projects  as  a  point,  and  cannot  be  made 
any  simpler. 


G 

L 

^<P< 

bh- 

r 

Fig.  90. 


VII,  §  70]     SECONDARY  PLAXES  OF  PROJECTION       49 

The  simplest  projection  of  a  straight  line  is  a  point.  This 
projection  can  be  obtained  by  a  single  secondary  plane  of  projec- 
tion only  when  the  given  line  is  parallel  to  one  of  the  original 
coordinate  planes.     Let  the  line  A  (Fig.  91)  be  parallel  to  H; 


Fig.  91. 


Fig.  92. 


then  if  GXLX  is  taken  perpendicular  to  Ah,  the  line  A  will  be 
seen  endwise,  and  the  projection  Af  will  become  a  point. 

Another  simple  and  useful  projection  of  a  straight  line  re- 
sults when  the  line  is  parallel  to  a  plane  of  projection.  This 
can  always  be  attained  by  means  of  a  secondary  F-plane,  by 
taking  GJjy  parallel  to  the  //-projection  of  the  line,  as  shown 
in  Figs.  88,  89,  and  90. 

The  simplest  position  of  a  plane  is  that  in  which  one  trace 
is  an  edge  view  of  the  plane.  Such  a  view  can  always  be  ob- 
tained by  means  of  a  secondary  vertical  plane  of  projection. 
Let  Q,  Fig.  92jbe  a  general  plane.  Assume  a  secondary  ground 
line  fr'Jjj  perpendicular  to  HQ.  Then,  since  IJQ  is  perpendicular 
to  this  ground  line,  Q  is  perpendicular  to  Vi  (§  50,  par.  5),  and 
its  trace,  ViQ,  on  the  secondary  F-plane  is  an  edge  view  of  Q 
(§  51),  To  find  ViQ:  since  Q  is  seen  edgewise  against  Vu  VXQ 
will  pass  through  fche  projection  of  any  point  in  Q. 

Now  VQ,  being  the  intersection  of  V  and  Q,  lies  in  both  V 
and  Q.  Assume  any  point  c  in  VQ;  then  the  poinl  c  is  in  Q 
and  also  in  V,  whence  c*  is  in  GL.  Using  the  projections  c* 
and  e",  find  the  secondary  projection  cf.     The  required  trace, 


50 


DESCRIPTIVE  GEOMETRY 


[VII,  §  70 


ViQ,  passes  through  c^  and  the  point  in  which  HQ  intersects 
the  ground  line  GlLi. 

71.  Projections  of  Solids.  A  secondary  projection  of  a  solid 
is  obtained  by  extending  the  method  used  for  points  and  lines. 
As  an  illustration,  the  irregular  triangular  pyramid  of  Fig.  32 
is  shown  in  Fig.  93  projected  on  a  secondary  vertical  plane.  In 
this  illustration  no  attempt  was  made  to  simplify  the  projec- 
tion of  the  solid,  nor  is  it  always  possible  to  do  so.  A  case  in 
which  a  simplification  is  effected,  however,  is  shown  in  Fig.  94  ; 


Fig.  m. 


Fn;.  94. 


here  the  projection  on  the  secondary  Fplane  gives  information 
about  the  size,  shape,  and  position  of  the  prism  which  is  not 
directly  obtainable  from  the  original  ^-projection. 

72.  The  Two  V-Projections  Compared.  The  relation  between 
two  ^-projections  of  an  object  is  the  same,  whichever  be  con- 
sidered the  original,  and  which  the  secondary,  projection  (§  67). 
Hence,  in  Fig.  94,  instead  of  constructing  the  Pi-projection 
from  the  //-  and  F-p rejections,  the  ^-projection  might  be  con- 
structed if  the  H-  and  p^-projections  are  known.  That  is.  if 
an  object  is  known  to  have  a  simple  projection  when  seen  from 


VII,  §  73]     SECONDARY  PLANES  OF  PROJECTION       51 

a  certain  point  of  view,  this  fact  may  be  used  to  construct  the 
projections  of  the  object.  The  student  should  examine  Fig.  94, 
assuming  it  to  be  drawn  as  the  answer  to  the  following  problem : 

Draw  the  H-  and  F-p rejections  of  a  square  right  prism. 
The  edges  of  the  base  are  f "  ;  length  of  prism  li".  The  lateral 
edges  are  parallel  to  H,  45°  with  V,  backward  to  the  left ;  the 
front  end  of  the  highest  lateral  edge  is  located  at  point 
1  (1A,  lv).    The  front  lateral  face  of  the  prism  makes  60°  with  H. 

[Hint.  After  point  1  is  located  in  all  three  projections,  the 
views  are  completed  in  this  order  :  IVpi'ojection,  ^-projection, 
F-projection.] 

The  object  shown  in  Fig.  95  consists  of  two  circular  conical 


Fin.  05. 


frustums  placed  base  to  base.     Note  the  relation  between  the  two 
^projections,  and  especially  the  position  of  the  ground  line. 

73.  Oblique  Secondary  Planes.  Secondary  planes  of  projec- 
tion perpendicular  to  Fand  oblique  to  II  will  not  be  considered 
in  detail,  as  their  use  in  practical  work  is  very  limited. 
(See  §  65.)  The  treatment  being  entirely  analogous  to  the 
plane  perpendicular  to  //,  the  student  should  have  no  difficulty 
in  using  such  a  plane  of  projection  should  occasion  arise. 


CHAPTER   VIII 

REVOLUTION  OF  A  POINT  — TRUE  LENGTH  OF  A 
STRAIGHT  LINE  —  APPLICATIONS 

74.  Revolution  of  a  Point  about  a  Straight  Line.  A  point  may 
be  revolved  about  any  straight  line  as  an  axis  through  a  com- 
plete revolution  or  through  a  portion  of  a  revolution.  In  the 
case  of  a  complete  revolution,  the  path  of  the  revolving  point 
will  be  the  circumference  of  a  circle  which  lies  in  a  plane  pass- 
ing through  the  point  and  perpendicular  to  the  axis  and  whose 


bv    a      o"  c 

I 

I  M  I 

I              I  I 

I              I  I 

/  j     I  X  i 

bl           i  Ic*1 


Fig.  96. 

center  lies  on  the  axis.  In  the  case  of  a  portion  of  a  revolu- 
tion, the  path  will  be  an  arc  of  such  a  circle. 

We  shall  proceed  to  discuss  certain  simple  cases  in  which  the 
axis  of  revolution  lies  in  a  particular  position. 

75.  Axes  of  Revolution  Perpendicular  to  if  or  V.  In  certain 
positions  of  the  axis  of  revolution,  the  circular  path  of  the 
revolving  point  will  have  simple  projections  on  either  H  or  V, 
or  both. 

Let  a,  Fig.  96,  be  revolved  about  an  axis  perpendicular  to 

52 


VIII,  §  76] 


REVOLUTION  OF  A  POINT 


53 


H.  The  circle  in  which  a  revolves  will  be  parallel  to  H,  and 
will  project  on  H  in  its  true  shape  and  size,  while  the  F-pro- 
jection  will  be  a  straight  line  parallel  to  the  ground  line. 
Similarly  (Fig.  97),  if  the  point  is  revolved  about  an  axis  per- 


ri  \>c  j 

!\ '  '  A 

i  i 

i  i 

1  •  ! 

b»i      I.  V  U 


Fig.  97. 


pendicular  to  V,  the  circular  path  projects  on   7as  a  circle, 
and  on  if  as  a  straight  line  parallel  to  GL. 

76.    Revolution  of  a  Point  about  an  Axis  Lying  in  a  Given  Plane 
(Fig.  98).     Let  us  consider  a  point  a,  which  is  to  be  revolved 


Fig.  98. 

about  an  n x is  X,  lying  in  a  plane  Q.  The  circular  path 
•  it  the  revolving  point  will  project  on  Q  as  a  straight  line,  be, 
perpendicular  to  the  axis  X.  It  follows  that  if  the  point  a 
be  revolved  into  the  plane  Q,  its  revolved  position,  b  or  c, 
will  be  at  a  distance  he  or  ce  from  the  axis  X,  equal  to  the 
true  distance  ae,  of  the  point  from  the  axis. 


54 


DESCRIPTIVE   GEOMETRY 


[VIII,  §  77 


77.  The  True  Length  of  a  Line.  Let  ab  be  a  straight  line  of 
definite  length.  If  placed  parallel  to  any  plane  of  projection, 
as  Q,  Fig.  99,  the  projection  of  aqbq  will  evidently  be  equal  hi 
length  to  ab.  If,  however,  the  line  is  placed  at  any  angle  with 
Q,  the  length  of  the  projection  will  be  shorter  than  that  of  the 
line.     (See  the  line  cd,  Fig.  39.)     But  by  revolution  about  a 


Fig.  99. 

suitable  axis,  a  line  can  always  be  brought  into  such  a  position 
that  the  true  length  will  appear. 

78.  First  Method  for  Finding  True  Length.  The  projection 
of  a  line  is  equal  to  its  true  length  when  the  line  is  parallel  to 
the  plane  of  projection  (§  77).  The  true  length  of  any  straight 
line  may  therefore  be  found  by  revolving  the  line  until  it  is 
parallel  to  either  H  or  V. 

Problem  3.     To  find  the  true  length  of  a  straight  line. 

First  Method.  By  revolving  about  an  axis  perpendicular 
to  H  or  V. 

Analysis.  Through  either  end  of  the  line  assume  an  axis 
perpendicular  to  H  and  (of  necessity)  parallel  to  V.  Revolve 
the  given  line  about  this  axis  until  the  line  is  parallel  to  V. 
Then  the  Tr-projection  of  the  line  in  its  revolved  position  will 
show  the  true  length  of  the  line. 

Or,  assume  the  axis  perpendicular  to  V  and  hence  parallel 
to  H,  and  revolve  the  given  line  until  parallel  to  H. 

Construction  (Fig.  100).  Let  ab  (akbh,  a'b")  be  the  given 
line.  Assume  an  axis  of  revolution  perpendicular  to  H  to 
pass  through  the  point  a.  Revolve  the  line  about  this  axis 
until  the  /^-projection  takes  the  position  a"ch,  parallel  to  OL. 


VIII,  §  79] 


TRUE  LENGTH  OF  A  LINE 


55 


In  this  revolution,  the  angle  which  the  line  makes  with  the 
axis  does  not  change.  Consequently,  the  angle  which  the 
line  makes  with  H,  and  therefore  the  length  of  the  if-projection, 
remain  constant.  The  point  b  revolves  in  the  circular  arc 
J(Jh,  Jv),  lying  in  a  plane  parallel  to  7/  (§  75),  and  is  found  at 
the  point  c  (ch,  c").  The  line  in  the  position  ac  (ahch,  a^c")  is 
parallel  to  V,  and  cfc"  shows  the  true  length. 


Fig.  100. 


Fig.  101. 


In  a  second  example,  Fig.  101,  the  axis  of  revolution  is 
taken  through  the  point  b  and  perpendicular  to  V.  The  true 
length  appears  after  the  line  has  been  revolved  parallel  to  //. 

79.  The  Angles  Which  a  Line  Makes  with  the  Coordinate  Planes. 
A  line  in  the  position  ac  (<<hch,  avcv),  Fig.  100,  which  is  parallel  to 
V,  shows  in  the  ^projection  the  angle,  a,  which  the  line  itself 
makes  with  //.  The  line  ac  is  the  revolved  position  of  the  line 
ab.  In  the  preceding  construction  we  saw  that  during  the  revo- 
lution the  angle  which  the  line  made  with  H  did  not  change. 
Hence  a  is  the  angle  which  the  given  line  ab  makes  with  II. 
That  is,  in  addition  to  finding  the  true  length  of  the  line  ab  in 
Fig.  100,  we  have  incidentally  found  the  angle  which  this  line 
makes  with  //. 

Similarly,  in  Fig.  Kll.  we  found,  ;is  an  incidental  part  of  the 
construction,  the  angle  (3  which  the  line  ab  makes  with    1'. 


56 


DESCRIPTIVE  GEOMETRY 


[VIII,  §  80 


80.  Second  Method  for  Finding  True  Length.  The  true  length 
of  a  straight  line  may  also  be  found  by  revolving  the  line  about 
one  of  its  own  projections  as  an  axis,  until  the  line  lies  in  a 
coordinate  plane. 

Problem  3  (bis).     To  find  the  true  length  of  a  straight  line. 

Second  Method.  By  revolution  about  one  of  the  projec- 
tions of  the  line. 


Fig.  102. 

Analysis  (Fig.  102).  Let  ab  be  the  given  line,  projected  on 
any  plane  Q  at  aqbq.  Revolve  the  plane  abbqa?  about  aqbq  as 
axis  into  Q.  Then  a  falls  at  ar,  where  ara"  equals  aa",  and  is 
perpendicular  to  aqb"  (§  76) ;  b  falls  at  br,  with  brbq  equal  to 
bbq  and  perpendicular  to  aqbq ;  hence  arbr  equals  ab,  and  shows 
the  true  length  of  the  line. 

Construction.  The  projection  ahbh,  Fig.  103,  corresponds  to 
aqbq,  Fig.  102,  and  is  taken  as  the  axis  of  revolution.  Since  a* 
is  the  projection  of  the  point  a  in  space,  the  distance,  aah,  of 
point  a  from  the  axis  is  equal  to  the  distance  of  the  point  a 
from  H,  which  shows  in  the  ^projection  as  the  distance  from  av 
to  GL.  Hence  to  find  the  revolved  position  ar,  make  ahar  per- 
pendicular to  ahbh,  and  the  distance  ahar  equal  to  ave.  Simi- 
larly, bhbr  is  perpendicular  to  ahbh,  and  is  equal  to  bvf.  Then 
arbr  shows  the  true  length  of  the  line  ab. 

In  Fig.  104  the  F-projection  is  taken  as  the  axis,  and  the 
line  is  revolved  into  V.  The  distances  avar  and  bvbr  are  the 
distances  of  the  points  a  and  b,  respectively,  from  V,  and  are 
equal  to  the  distances  from  ah  and  bh  to  the  ground  line.  !NTote 
that  not  only  the  distances,  but  the  directions,  of  the  points  a 
and  b  from  the  F-plane  must  be  considered.    The  //-projection 


VIII,  §  82] 


TRUE  LENGTH  OF  A  LINE 


57 


shows  that  a  and  b,  in  space,  are  on  opposite  sides  of  V,  hence 
their  revolved  positions,  ar  and  br,  must  fall  on  opposite  sides 
of  the  axis,  avb",  which  lies  in  V. 


Fig.  103. 


Fig.  104. 


81.  Traces  of  the  Line.  Let  the  line  ab,  Fig.  102,  be  pro- 
duced to  intersect  its  projection  aqb"  in  the  point  s ;  then  in 
revolving  about  aqbq  as  axis,  the  point  s  will  remain  fixed,  and 
arbT  produced  must  pass  through  s.  But  since  a"b"  lies  in  Q, 
the  point  s  must  be  the  point  in  which  ab  pierces  Q,  or  the 
Q-trace  of  ab.  In  the  projection,  Fig.  103,  let  arbr  be  produced 
to  meet  ahbh  produced  at  s ;  then  8  must  be  the  //-trace  of  the 
line  ab,  and  should  be  the  same  point  that  is  found  by  project- 
ing from  the  intersection  of  avbv  produced  and  GL,  as  in 
Problem  1  (§  37). 

82.  Angles  with  the  Coordinate  Planes.  The  angle  a,  Fig. 
102,  between  ab  and  a''//'  is  the  angle  which  the  line  ab  makes 
with  the  plane  Q,  and  this  angle  is  not  changed  by  the  revo- 
lution into  Q.  Hence,  in  Fig.  103,  the  angle  a,  between  ahbh 
and  a,.br,  is  the  true  angle  which  the  line  ab  makes  with  //. 

In  Fig.  104,  since  the  line  ab  was  revolved  into  V,  we  have 
found  the  F-trace,  t,  where  arbr  intersects  avbv.  The  angle,  (3, 
between  these  lines  is  the  angle  which  the  line  ab  mukes  with  V. 


58 


DESCRIPTIVE  GEOMETRY 


[VIII,  §  83 


83.  Converse  Problems.  In  each  of  the  preceding  methods 
for  finding  the  true  length  of  a  line  it  is  necessary  to  find  the 
angle  between  the  line  and  one  of  the  coordinate  planes. 
Conversely,  it  is  easy  to  see  that  a  line  may  be  located,  when 
certain  angles  are  known,  by  reversing  one  of  the  preceding 
constructions. 

Problem  4.  To  find  the  projections  of  a  line  of  definite  length, 
when  its  slope,  the  angle  which  it  makes  with  one  coordinate  plane, 
and  the  direction  of  its  projection  on  that  plane,  are  known. 

In  order  to  give  the  line  a  definite  position  in  space,  it  will 
be  assumed,  in  addition  to  the  above  data,  that  one  end  of  the 
line  is  located  at  a  given  fixed  point  a. 

It  is  evident  geometrically  that  the  position  of  the  line  in 
space  is  now  completely  determined. 

Analysis.  If  the  angle  which  the  line  makes  with  H  is 
given,  reverse  any  construction  of  Problem  3  which  gives  the 
angle  with  H.  Proceed  similarly  if  the  angle  with  V  is  the 
given  angle. 


Fig.  100  (repeated). 


Fig.  101  (repeated). 


Construction.  First  Method  (Fig.  100).  [Reverse  of 
First  Method,  Prob.  3,  §  78.]  Let  the  angle  with  H  and 
the  direction  of  the  if-projection  be  given.  Let  the  point  a  be 
the  fixed  end  of  the  line. 


VIII,  §  83] 


TRUE  LENGTH  OF  A  LINE 


59 


Place  the  line  in  the  position  ac  (avcv,  ahch),  parallel  to  V, 
making  avcv  equal  to  the  true  length  of  the  line,  at  the  given 
angle  with  H,  and  sloping  upward  or  downward  from  a  as 
given. 

Take  an  axis  perpendicular  to  H  through  the  fixed  end,  a, 
and  revolve  the  line  until  the  //-projection  takes  the  given 
direction  ahbh.     Find  the  corresponding  projection  avb". 

Second  Method  (Fig.  103).  [Reverse  of  Second  Method, 
Prob.  3,  §  80.]     As  before,  let  the  angle,  a,  with  II  and  the 


-A- 


/ 


\a    L 


X 


Fig.  103  (repeated). 


Fig.  104  (repeated). 


direction  of  the  H"-projection  be  given,  and  let  the  point  a  be 
the  fixed  end  of  the  line.  Through  a'1  draw  ahs,  indefinite  in 
length,  and  in  the  given  direction.  Draw  ahar  perpendicular 
to  ahs,  and  make  the  distance  ahar  equal  to  ave.  Draw  ars  in- 
definite in  length  and  at  the  angle  a  with  ahs.  In  this  case,  the 
line  is  supposed  to  slope  downward  from  the  point  a.  Make, 
arbr  equal  to  the  given  true  length  of  the  line.  Draw  brbh  per- 
pendicular to  ats,  locating  bh.  Project  from  &*  perpendicular 
to  GL,  and  make  the  distance  fbv  equal  to  bhbr,  thus  locating  &". 

The  student  should  see  if  he  can  solve  the  problem  when 
the  angle,  with  P"and  direction  of  the  ^projection  are  given, 
by  reversing  the  constructions  of  Figs.  101  and  104. 


60 


DESCRIPTIVE  GEOMETRY 


[VIII,  §  83 


Problem  5.  To  find  the  projections  of  a  line  making  given  angles 
with  H  and  V. 

In  order  to  bring  a  definite  solution  it  will  be  assumed  that 
the  true  length,  slope,  and  location  of  one  end  of  the  line  are 
known.  The  general  solution  to  the  problem  as  stated  above 
consists  of  four  series  of  parallel  lines,  corresponding  to  the 
four  possible  slopes  of  a  line  (§  31). 

Analysis.  Consider  the  problem  solved,  as  in  Fig.  105,  and 
let  ab  (ahbh,  avbv)  be  the  resulting  line,  with  point  a  the  fixed 


Fig.  105. 

end.  Take  an  axis  through  a  perpendicular  to  H,  and  revolve 
the  line  to  the  position  ac  (ahch,  a"cv),  parallel  to  V,  where  aV 
shows  the  true  length  of  the  line  and  its  angle,  a,  with  H 
(Prob.  3,  First  Method,  §  78).  Also  take  an  axis  through  a 
perpendicular  to  V,  and  revolve  the  line  to  the  position 
ad  (avclv,  ahdh)  parallel  to  H,  where  ahdh  shows  the  true  length  of 
the  line  and  the  angle,  /?,  with  V.  The  problem  may  now  be 
solved  by  reversing  this  construction. 

Construction.  Place  the  line  in  the  position  ac  (avcv,  ahch)  par- 
allel to  Y,  where  avcv  is  equal  to  the  true  length  of  the  line  and 
makes  with  GL  the  angle,  a,  which  the  line  makes  with  H. 


VIII,  §  83]  TRUE  LENGTH  OF  A  LINE  61 

Revolve  this  line  about  an  axis  through  a  perpendicular  to  H, 
by  drawing  the  projections,  Jh  and  Jv,  of  the  arc  in  which 
point  c  revolves. 

Place  the  line  also  in  the  position  ad  (ahdh,  avd")  parallel  to 
H,  where  ahdh  (equals  avcv)  equals  the  true  length  of  the  line 
and  makes  with  GL  the  angle,  /?,  which  the  line  makes  with 
V.  Revolve  this  line  about  an  axis  through  a  perpendicular  to 
V,  by  drawing  the  projections,  K"  and  Kh,  of  the  arc  in  which 
d  revolves. 

The  point  b  (bh,  bv),  which  determines  the  projections  of  the 
required  line  ab,  is  found  at  the  intersection  of  the  two  paths 
of  revolution  J  and  A".  Note  that  this  point  is  independently 
determined  in  each  projection;  hence  as  a  check  on  the  work 
the  projector  bvbh  should  be  perpendicular  to  GL. 


CHAPTER   IX 
SOME    SIMPLE    INTERSECTIONS  —  DEVELOPMENTS 

84.  Simple  Intersections  of  Solids  of  Revolution.  In  generat- 
ing the  surface  of  a  solid  of  revolution,  each  point  of  the  gener- 
ating line  describes  a  circle  which  lies  in  a  plane  perpendicular 
to  the  axis,  and  whose  center  lies  in  the  axis  (§  74).  If  the 
axis  of  the  surface  be  placed  perpendicular  to  H  (or  V),  these 
circles  will  project  on  H  (V)  in  their  true  shape  and  size,  and 
on  V  (H)  as  horizontal  straight  lines  which  are  the  edge  views 
of  the  planes  containing  the  circles  (§  75). 

The  intersection  of  a  surface  of  revolution  by  a  plane  may 
be  found  by  drawing  on  the  surface  a  sufficient  number  of 
circles,  obtained  by  passing  auxiliary  planes  perpendicular  to 
the  axis  of  the  surface,  and  then  finding  the  points  in  which 
these  circles  are  intersected  by  the  given  planes. 

85.  Visibility  of  the  Intersection.  In  this  and  subsequent 
parts  of  this  book,  whenever  a  solid  is  intersected  by  a  plane, 
no  part  of  the  solid  will  be  considered  as  cut  away  or  re- 
moved unless  so  stated.  In  determining  the  visibility  of  the 
line  of  intersection,  the  cutting  plane  will  be  considered  trans- 
parent and  the  solid  opaque.  The  line  of  intersection  will  be 
visualized  as  a  line  drawn  on  the  surface  of  the  solid.  Hence, 
the  line  of  intersection,  or  any  portion  of  it,  will  be  visible  in 
any  given  projection  when  the  line  lies  on  a  portion  of  the  sur- 
face which  is  visible  in  that  projection ;  and  conversely. 

The  visibility  of  the  line  of  intersection,  then,  depends 
directly  upon  the  visibility  of  the  given  solid  (§  24). 

86.  Examples.  The  only  position  of  the  cutting  plane 
which  will  be  considered  at  this  time  is  an  edge  view.     Then 

62 


IX,  §  86]     INTERSECTIONS  —  DEVELOPMENTS 


63 


the  plane  is  perpendicular  to  either  H  or  V.  In  the  following 
examples,  the  axis  of  each  surface  is  perpendicular  to  H,  so 
that  the  auxiliary  circles  (§  84)  all  lie  in  planes  which  are 
parallel  to  H.  Note,  in  each  example  the  visibility  of  the 
curve  of  intersection  (§  85). 

Example  1  (Fig.  106).  A  cone  of  revolution  cut  byaplane 
perpendicular  to  V.  Since  the  whole  convex  surface  of  the 
cone  is  visible  in  plan,  the  entire  curve  is  visible  in  plan. 


Fig.  li ii J 


Fig.  107. 


The  true  size  of  the  section  is  found  by  revolving  the  section 
until  parallel  to  V  about  an  axis  parallel  to  VQ.  The  true 
width  at  any  point,  as  2'8',  equals  the  true  distance  2-8  which 
appears  in  the  fl"-projection. 

Example  -  (Fig.  107).  A  roue  of  revolution  cut  by  a  plane 
parallel  to  V.  The  curve  is  visible  in  elevation,  since  it  ia 
wholly  on   the   front  half  of  the  cone. 

The  true  si/e  of  the  .section  appears  in  the  ^projection. 


64 


DESCRIPTIVE   GEOMETRY 


[IX,  §  86 


Example  3.  (Eig.  108).  A  sphere  cut  by  a  plane  perpen- 
dicular to  V.  Here  the  intersection,  as  seen  in  elevation,  lies 
partly  on  the  upper  half  of  the  sphere,  and  partly  on  the 
lower.  Hence,  in  the  plan  one  part  of  the  curve  is  visible,  and 
the  other  part  invisible.  The  two  points  in  which  the  curve 
changes  from  visible  to  invisible  are  necessarily  on  the  outline 
of  the  sphere.     Why  ? 

The  true  size  of  the  section  is  a  circle  whose  diameter 
appears  in  the  F-projection. 


Fig.  108. 


Fig.  109. 


Example  4  (Fig.  109).  A  torus  (§  26)  cut  by  a  plane  per- 
pendicular to  V.  An  auxiliary  plane,  such  as  M,  cuts  the  gen- 
erating circle  in  two  points.  Hence,  in  this  plane  two  circles 
lying  in  the  surface  of  the  torus  can  be  drawn,  one  of  diameter 
ad,  and  the  other  of  diameter  be.  Similarly,  two  circles  can 
be  drawn  in  every  plane  perpendicular  to  the  axis,  except  in 


IX,  §  86]     INTERSECTIONS  —  DEVELOPMENTS 


65 


those  which  pass  through  the  highest  and  lowest  points  of  the 
generating  circles. 

The  visibility  of  the  curve  in  plan  is  determined  in  the  same 
way  as  for  the  sphere,  Fig.  108. 

The  true  size  of  the  section  is  shown  parallel  to  V,  as  in 
Example  1. 

Example  5  (Fig.  110).  A  hyperbolic  spindle,  cut  by  a  plane 
perpendicular  to  V.  In  this  case  the  surface  that  is  cut  is 
wholly  hidden  in  plan,  hence  the  curve  is  invisible. 

The  true  size  of  the  section  is  shown  parallel  to  V. 


Fig.  110. 


Fig.  ill. 


Example  6  (Fig.  111).  A  torus,  cut  by  two  planes  parallel 
to  V.  The  X  section,  wholly  on  the  hack  of  the  surface,  is 
invisible.  The  FTsection,  entirely  on  the  visible  part  of  the 
torus,  is  wholly  visible. 

The  true,  size  of  each  section  appears  at  once. 


66 


DESCRIPTIVE  GEOMETRY 


[IX,  §  86 


Example  7  (Fig.  112).      A   hyperbolic   spindle,    cut   by  tv:o 
planes  parallel  to  V.     Each  plane  cuts  a  section  composed  of 


Fig.  112. 


two  parts.     The  visibility  of  the  curves  should  be  evident  from 
an  inspection  of  the  position  of  the  cutting  planes. 

87.  Developments.  The  development  of  a  solid  bounded  by 
plane  faces  consists  of  a  series  of  polygons  of  the  true  sizes 
and  shapes  of  the  various  faces  of  the  solid,  so  arranged,  edge 
to  edge,  that  the  development  may  be  folded  up  to  reproduce 
the  surface  of  the  solid.  If  a  face  of  a  solid  is  a  triangle,  the 
true  size  of  the  face  may  be  constructed  from  the  true  lengths 
of  its  three  sides.  If  the  face  is  a  polygon  of  more  than  three 
sides,  the  true  size  and  shape  cannot  be  determined  from  the 
true  lengths  of  the  sides  alone.  Any  plane  polygon,  however, 
can  be  divided  into  triangles  by  drawing  certain  diagonals,  and 
the  polygon  can  be  constructed  from  these  triangles.  Hence 
the  development  of  any  solid  bounded  by  plane  faces  can  be 
obtained  wholly  by  finding  the  true  lengths  of  straight  lines. 


IX,  §  88]     INTERSECTIONS  —  DEVELOPMENTS 


67 


88.    Working  Method  for  Finding  the  True  Length  of  a  Line. 

In  the  solution  of  many  problems  the  true  length  of  a  line  is 
found  in  order  to  be  used  as  a  radius  for  setting  compasses  or 
dividers.  The  angle  made  with  H  or  V  is  not  needed.  Hence 
it  is  sufficient  to  construct  two  points  whose  distance  apart  is 
equal  to  the  true  length  of  the  line. 

Let  ab,  Fig.  113,  be  any  line,  and  let  the  true  length  of  this 
line  be  found  by  revolving  it  parallel  to  V,  keeping  the  upper 


Fig.  113 


end  of  the  line  fixed  (Prob.  3,  First  Method,  §  78),  thus  giving 
avc  as  the  true  length.  Now  the  point  c  can  be  found  by  draw- 
ing through  bv  a  horizontal  line,  intersecting  the  projector  avah 
at  e,  and  then  laying  off  the  distance  ec  directly  with  the 
dividers  ecpial  to  ukbh.     Hence  the  following  working  rule : 

Rule.  Tlie  true  length  of  a  straight  line  equals  the  hypothe- 
nuse  of  a  right  triangle  whose  base  equals  the  length  of  the  H-pro- 
jection  of  the  line,  and  whose  altitude  in  equal  to  the  difference  in 
elevation  between  the  two  ends  of  the  line. 

If  the  lower  end  of  the  line  is  in  the  HT-plane,  6*  will  lie  in  OL, 
and  the  method  becomes  even  simpler.  Thus,  the  true  length 
of  the  line  ab,  Fig.  114,  may  be  found  by  laving  off  on  OL  the 
distance  ec  equal  to  ahbh;  then  avc  is  the  desired  true  length. 


68 


DESCRIPTIVE  GEOMETRY 


[IX,  §  89 


89.  To  Find  the  Development  of  a  Solid  with  Plane  Faces. 
Let  the  given  solid  be  the  frustum  of  an  irregular  four-sided 
pyramid,  Fig.  115.  Each  face  is  a  quadrilateral,  which  can  be 
divided  into  two  triangles  by  means  of  one  diagonal.  Never- 
theless it  is  easier  to  begin  by  finding  the  development  of  the 
triangular  faces  of  the  complete  pyramid.  Assuming  that  the 
true  lengths  of  the  various  lines  are  found  as  needed,  we  proceed 
with  them  as  follows  (B,  Fig.  115).     On  any  convenient  line, 


Fig.  115. 


lay  off  the  distance  0-1 ;  with  0  and  1  as  centers,  and  radii  0-2 
and  1-2  respectively,  strike  arcs  intersecting  in  point  2.  This 
gives  the  development  of  the  triangular  face  0-1-2. 

With  0  and  2  as  centers,  radii  0-3  and  2-3  respectively, 
locate  point  3,  thus  obtaining  the  development  of  the  face 
0-2-3.     In  the  same  way  obtain  the  faces  0-3^  and  0-4-1. 

The  base,  1-2-3-4,  of  the  pyramid  is  divided  into  triangles 
by  the  diagonal  1-3,  and  is  plotted  to  join  one  of  the  edges 
previously  located,  as  2-3.  The  result  thus  far  obtained  is  the 
development  of  the  complete  pyramid.  To  convert  it  into  the 
development  of  the  frustum,  proceed  as  follows  : 


IX,  §  90]     INTERSECTIONS  —  DEVELOPMENTS 


69 


On  0-1,  0-2,  etc.,  of  the  development,  measure  the  true 
lengths  of  the  lines  1-5,  2-6,  3-7,  4-8,  and  draw  5-6-7-8-5. 

Draw  one  diagonal,  as  5-7,  of  the  upper  base,  plot  this  base 
by  means  of  the  two  triangles  thus  formed,  and  join  it  prop- 
erly to  one  of  the  previous  lines  of  the  development.  The 
development  of  the  frustum  is  now  complete. 

90.  To  Construct  the  Projections  of  a  Prism  Whose  Long  Edges 
Make  Given  Angles  with  fl"and  V  (Fig.  116).  Let  it  be  required 
to  draw  the  projections  of  a  prism  whose  long  edges  shall  make 
30°  with  H  and  45°  with  V. 

Place  the  prism  first  with  its  long  edges  parallel  to  H  and 
making  the  given  angle,  45°,  with  P".  Assuming  that  the  neces- 
sary sizes,  slopes,  and  distances  are  given,  we  can  construct 


Fig.  ll«. 


these  projections  by  using  a  secondary  ground  line,  as  in  §  72 
(B,  Fig.  116).  Find,  by  Problem  5,  §  83,  the  projections  of  a 
line  ab  making  the  given  angles  with  II  and  V  (A,  Fig.  116). 

Now  the  long  edges  of  the  prism  as  placed  in  B,  Fig.  116, 
make  the  given  angle  with  V.  If  the  prism  be  revolved  about 
any  axis  perpendicular  to  V,  the  edges  will  continue  to  make 
the  same  angle  with  V,  while  the  projection  of  the  prism  on  V 


70 


DESCRIPTIVE   GEOMETRY 


[IX,  §  90 


will  remain  of  the  same  shape  and  size.  Let  the  prism  be  re- 
volved until  the  F-projections  of  the  long  edges  are  parallel  to 
avbv,  as  shown  at  C,  Fig.  116. 

In  making  this  revolution,  the  actual  axis  need  not  be  used. 
It  suffices  to  copy  the  F-projection  obtained  at  B,  making  the 
long  edges  of  the  prism  parallel  to  the  direction  avbv  obtained 
at  A.  But  in  this  revolution,  the  ^-projection  of  the  path  of 
each  moving  point  will  be  a  horizontal  straight  line,  regardless 


Fig.  116  (repeated). 

of  the  position  of  the  axis  of  revolution.  Hence  the  //"-projec- 
tion at  C  may  be  obtained  by  projecting  vertically  from  the 
F-projection,  and  horizontally  from  the  //-projection  at  B. 

As  a  check,  the  long  edges  of  the  projection  thus  obtained 
should  be  parallel  to  the  direction  ahbh  found  at  A. 

The  projections  at  C  are  the  required  projections. 

91.  Other  Methods.  The  projections  at  B,  Fig.  116,  were 
made  by  knowing  the  angle  which  the  long  edges  of  the  prism 
made  with  V,  while  the  projections  at  C  were  made  by  know- 
ing the  inclination  of  the  F-projection  of  the  edges.  These 
two  quantities  might  have  been  given  directly,  and  projections 
B  and  C  constructed  by  Problem  4,  §  83. 


CHAPTER   X 


LINES   IN  A   PLANE  —  PARALLEL  LINES   AND   PLANES 


92.  Intersecting  and  Parallel  Lines.  Two  lines  intersect 
when  they  pass  through  a  common  point.  Two  lines  are 
parallel  when  they  are  everywhere  equally  distant.  If  two 
straight  lines  lie  in  the  same  plane,  they  either  intersect  or  are 
parallel.  This  is  not  necessarily  the  case,  however,  with  any 
two  lines  in  space. 

93.  Parallel  Lines  in  Space.  If  two  lines  in  space  are  parallel, 
their  projections  on  any  plane  are  parallel. 

Let  the  two  parallel  lines  A  and  B,  Fig.  117,  be  projected 
on  the  plane  Q  by  planes  perpendicular  to  Q.     Since  the  two 


Ftg.   117. 


projecting  planes  will  be  parallel,  their  intersections,  Aq  and 
Bq,  with  the  plane  Q  will  be  parallel. 

In  general,  if  lines  A  and  l>  are  parallel,  it  is  definitely 
shown  by  the  simultaneous  condition  that  Ah  is  parallel  to  Ilh, 
and  A"  is  parallel  to  />'".  Profile  lines  are  an  exception  ;  such 
lines  may  be  tested  by  means  of  their  profile  projections. 
Thus,  the  profile  lines  ad  and  cd,  Fig.  IIS,  are  parallel,  since 
their  profile  projections  are  parallel. 

71 


72 


DESCRIPTIVE  GEOMETRY 


[X,  §  94 


94.  Intersecting  Lines  in  Space.  If  two  lines  in  space  intersect, 
their  corresponding  projections  intersect  in  the  same  projector. 

Let  A  and  B,  Fig.  119,  be  two  intersecting  lines.  Since 
these  lines  intersect,  there  must  be  a  point,  c,  common  to  each 
of  them.     Since  c  is  a  point  of  the  line  A,  the  Zf-p  rejection,  ch, 


Fig.  119. 

must  lie  on  Ah  (§  36).  Moreover,  since  c  is  in  the  line  B,  c* 
must  lie  on  Bh.     Hence  c*  is  at  the  intersection  of  Ah  and  B\ 

Similarly,  the  F-projection  of  c,  c",  must  lie  at  the  inter- 
section of  Av  and  B".  But  since  ch  and  c"  are  two  projections 
of  the  same  point,  they  must  lie  in  the  same  projector. 

If  two  lines  have  a  common  projection  (Fig.  120),  they  lie  in 
the  same  plane,  namely,  the  plane  whose  edge  view  coincides 


Fig.  120. 


with  the  common  projection.     Such  lines  are  either  intersect- 
ing, as  in  the  figure,  or  parallel  (§  92). 


X,  §  95] 


LINES  IN  A  PLANE 


73 


95.  Test  for  Intersecting  Lines.  Conversely,  if  the  correspond- 
ing projections  of  two  lines  intersect  in  the  same  projector,  the 
lines  in  space  intersect.     The  only  exception  occurs  when  one  of 


Fig.  121. 


the  lines  is  a  profile  line ;  thus,  the  line  E,  Fig.  121,  does  not 
intersect  the  profile  line  ab,  since  the  point  c,  in  which  E  pierces 
the  profile  plane  containing  ab,  does  not  lie  on  the  line  ab. 


If  the  point  of  intersection  is  outside  the  limits  of  the  draw- 
ing (Fig.  122),  the  test  fails.  In  this  case  take  any  two  points, 
d  and  e,  on  A,  and  any  two,/  and  g,  on  B.  Then  if  A  and  B 
intersect,  df  and  eg  (or  else  dg  and  ef)  will  intersect. 


74 


DESCRIPTIVE  GEOMETRY 


[X,  §  96 


96.  A  Line  in  a  Plane.  If  a  line  lies  in  a  plane,  the  traces 
of  the  line  lie  in  the  corresponding  traces  of  the  plane. 

Let  a  line  A  lie  in  a  plane  Q  (Figs.  123  and  124).  Then  if 
A  intersects  H,  it  can  evidently  do  so  only  in  some  point  of 


Fig.  123. 


Fig.  124. 


the  line  in  which  Q  intersects  H ;  and  the  intersections  with 
H  are  the  respective  traces  of  the  line  and  plane.  Similarly 
for  any  other  plane  of  projection. 

97.  To  Locate  a  Line  in  a  Plane.  A  line  will  lie  in  a  given 
plane  if  the  line  passes  through  two  points  which  lie  in  the 
plane.  Thus,  in  Fig.  123,  let  us  assume  sh  on  HQ,  and  tv  on  VQ. 
These  two  points  are  the  traces  of  some  line  lying  in  the  plane  Q 
(§  96).  The  projections  Ah  and  A"  may  then  be  found  by  Prob- 
lem 2,  §  37. 

Other  methods  of  determiniDg  a  line  so  that  it  shall  lie  in 
a  given  plane  will  be  given  later.    (See  §§  100,  133.) 

98.  A  Plane  Containing  a  Given  Line.  The  proposition  of 
§  96  enables  us  to  pass  a  plane  through  a  given  line.  Thus,  in 
Fig.  123,  let  A(Ah,  A")  be  a  given  line,  whose  traces  are  the 
points  s  and  t.  A  plane  Q  may  be  passed  through  A  by 
drawing  HQ  in  any  direction  through  s*,  and  then  drawing 
VQ  through  tv  and  the  point  in  which  HQ  intersects  the 
ground  line.  Conversely,  VQ  may  be  drawn  first,  in  any  direc- 
tion through  f,  and  then  HQ  may  be  drawn  through  sh  and 
the  intersection  of  VQ  with  the  ground  line. 

An   indefinite  number  of  planes  may  be  passed  through  a 


X,  §  99] 


LINES  IX  A  PLANE 


75 


given  line.  In  particular  (Fig.  125),  we  may  draw  a  plane  X, 
perpendicular  to  H,  a  plane  Y  perpendicular  to  V,  and  a  plane 
S  parallel  to  the  ground  line.     (See  also  §  102.) 


Fig.  125. 


99.  Principal  Lines  of  a  Plane.  For  every  plane  there  are 
two  systems  of  lines  which  are  of  importance,  and  are  called 
the  principal  lines  of  the  plane.  One  of  these  systems  is 
parallel  to  II,  the  other  parallel  to  F. 

The  horizontal  principal  lines  are  //-parallels,  cut  from  the 
given  plane  by  a  series  of  planes  parallel  to  //.  The  lines  arc 
therefore  parallel  to  the  //-trace  of 
the  plane  ;  their  //-projections  are 
parallel  to  this  trace,  while  their 
^projections  are  parallel  to  the 
ground  line.  Visualize  this  state- 
ment, aided,  if  necessary,  by  Fig.  126. 

The  vertical  principal  lines  are  cut 
from  the  plane  by  a  series  of  planes 
parallel  to  V.  They  are  F-paralhls, 
parallel  to  the  F-trace  of  the  plane.  Their  F-projections  are 
parallel  to  this  trace,  their  ff-projections  are  parallel  to  the 
ground  line. 

These  tun  groups  of  lines  are,  in  general,  different.  They 
reduce  to  the  same  set  when,  and  onl\  when,  the  given  plane 
is  parallel  to  the  ground  line. 


/-^Av\ 

V 

/ 

7 

Fig.  126. 


76 


DESCRIPTIVE  GEOMETRY 


[X,  §  100 


100.    To  Project  the  Principal  Lines  of  a  General  Plane.     To 

draw  the  projections  of  a  horizontal  principal  line  of  a  general 
plane,  Q,  Fig.  127,  assume  the  F-trace  of  this  line  as  any  point, 


Fig.  127. 


t",  in  VQ,  and  find  th  on  GL.  Then,  by  §  99,  Ah  passes  through 
th  and  is  parallel  to  HQ,  while  Av  passes  through  f  and  is 
parallel  to  GL. 

Similarly,  a  vertical  principal  line,  B,  may  be  drawn  by  first 
assuming  its  ZZ-trace,  s,  on  HQ. 

101.  To  Project  the  Principal  Lines  of  a  Plane  Parallel  to 
the  Ground  Line.  To  draw  a  principal  line  of  a  plane  when 
the  plane  is  parallel  to  the  ground  line,  as  Q,  Fig.  128,  find  the 


Q. 
> 

VQ 

\fl 

uv               Av 

/     1 

/      1 

[        \ 
\ 

Uh            Ah 

HQ 

D. 

X 

Fig.  12S. 


profile  trace,  PQ,  of  the  given  plane.  On  PQ  assume  any 
point,  up,  as  the  profile  trace  of  the  required  principal  line. 
Both  projections  of  this  line  are  parallel  to  the  ground  line. 


X,  §  103]        PARALLEL  LINES  AND  PLANES 


77 


102.  A  Plane  Containing  a  Line  Parallel  to  H  or  V.  Any 
plane,  as  Q  or  R,  Fig.  129,  which  is  passed  through  an  H- 
parallel  will  have  its  //"-trace  parallel  to  the  //"-projection  of 
the  line.     (Converse  of  §  100.) 

Similarly,  any  plane  passed  through  a  F-parallel  will  have 
its  F-trace  parallel  to  the  F-projection  of  the  line. 

Any  plane,  as  Q,  Fig.  128,  which  contains  a  line  parallel  to 
both  H  arid  V  will  have  both  its  H-  and  F-traces  parallel  to 
the  ground  line.     (Converse  of  §  101.) 

103.  Parallel  Planes.  If  two  planes  are  parallel,  their  corre- 
sponding traces  are  parallel. 

Let  Q  and  R  be  the  given  planes.  Let  Q  intersect  H  in 
HQ,  and  R  intersect  //  in  HR.  The  figure  is  left  for  the 
student  to  draw.  Now  when  two  parallel  planes,  Q  and  R, 
are  intersected  by  a  third  plane,  H,  the  lines  of  intersection 
must  be  parallel.  Hence  HQ  is  parallel  to  HR.  Similarly 
for  any  other  plane  of  projection. 

VQ. 


7 

> 

HR 

N 

\ 
\ 

.  \ 
\\ 

a. 

X 

A 

/          VR 

HQ 

VS 

\ 

-HS 

Fia.  130. 


Conversely,  if  the  //-  and  F-traces  of  two  planes  are  parallel, 
the  planes  are,  in  general,  parallel.  The  only  exception  occurs 
with  planes  parallel  to  the  ground  line.  Such  planes  may  be 
tested  by  means  of  their  profile  traces.  Thus,  in  Fig.  130,  the 
plane  R  is  parallel  to  the  plane  Q,  while  the  plane  S  is  not. 


78 


DESCRIPTIVE   GEOMETRY 


[X,  §  104 


104.  A  Line  Parallel  to  a  Plane.  A  straight  line  is  parallel 
to  a  plane  if  it  is  parallel  to  some  line  lying  in  the  plane. 
Thus  the  line  A,  Fig.  131,  is  parallel  to  the  plane  Q,  since  it  is 
parallel  to  the  line  B  which  lies  in  Q.     If  the  line  B  were  not 


Fig.  131. 


drawn,  however,  it  would  be  difficult  to  recognize  the  parallel- 
ism. Thus,  in  Fig.  132,  both  the  lines  A  and  B  are  parallel  to 
the  plane  B,  but  this  fact  can  hardly  be  seen  by  inspection. 


Fig.  132. 


Fig.  133. 


The  fact  that  a  line  is  parallel  to  a  plane  can  be  recognized  at 
once  only  when  the  line  is  parallel  to  one  of  the  systems  of 
principal  lines  of  the  plane.  In  Fig.  133,  it  is  evident  that 
both  lines  C  and  D  are  parallel  to  the  plane  S. 


X,  §  106]        PARALLEL  LINES  AND  PLANES  79 

105.  A  Plane  Parallel  to  a  Line.  Conversely,  a  plane  is  par- 
allel to  a  straight  line  if  the  plane  contains  a  line  which  is 
parallel  to  the  given  line.  Thus,  in  Fig.  131,  since  the  lines  A 
and  B  are  parallel,  the  plane  Q,  or  any  other  plane  passed 
through  B,  is  parallel  to  the  line  A.  The  only  exception  is 
the  plane  which  contains  both  A  and  B. 

106.  A  Plane  Determined  by  Lines  or  Points.  One  plane,  and 
one  only,  may  be  found  which  contains 

(a)   two  intersecting  lines  ; 
(6)    two  parallel  lines  ; 

(c)  a  line  and  a  point  not  on  the  line ; 

(d)  three  points  not  in  the  same  straight  line. 

Problem  6.  To  find  the  plane  which  contains  two  given  intersecting 
or  parallel  lines. 


/  y\\ 

\^4 

^^P^ 

1 
1 

1 

V^ 

Fig.  134. 

Analysis.  The  trace  of  the  required  plane  on  any  coordinate 
plane  must  contain  the  trace  of  each  of  the  given  lines  (§  96). 
The  plane  is  found  when  its  II  and  V  traces  are  found. 

Construction  (Fig.  134).  Let  A  and  B,  intersecting  at  point 
c,  be  the  given  lines.  Find  the  traces  of  A  and  B  (Prob.  1, 
§  37).  The  required  trace  HQ  is  determined  by  the  two 
iZ-traces  sx  and  s2 ;  the  trace  VQ  by  the  two  P"-traces  tx  and  L. 

Check.  The  H-  and  F-traces  of  a  plane  must  intersect  on 
the  ground  line  (§  46).  Hence  II Q  and  VQ,  when  produced, 
must  intersect  on  GL. 


80 


DESCRIPTIVE  GEOMETRY 


[X,  §  106 


A  plane  which  contains  two  given  parallel  lines  is  shown  in 
Eig.  135.    The  construction  is  entirely  similar  to  that  just  given. 


Fig.  135. 


In  Fig.  136  the  IZ-trace  of  the  given  line  B  is  too  far  removed 
to  be  used  in  the  construction.     The  plane  Q,  containing  A  and 


Fig.  136.  Fig.  137. 

B,  is  located  by  first  drawing  VQ  through  the  F-traces  of  the 
given  lines ;  then  HQ  is  drawn  through  the  JT-trace  of  A  and 


X,  §  106]        PARALLEL  LINES  AND  PLAXES 


81 


the  point  in  which  VQ  intersects  the  ground  line.  This  con- 
struction gives  no  check  on  the  work,  but  may  be  used  in  case 
of  necessity.  Other  constructions  which  may  be  employed 
wheu  traces  of  the  given  lines  are  not  accessible  will  be  given 
in  §  108.  We  shall  consider  at  present  only  those  cases  in 
which  one  or  both  of  the  given  lines  are  parallel  to  H  or  to  V. 

Special  Case  I.  Let  us  suppose  that  one  of  the  given  lines 
is  parallel  to  H  or  V.  Let  A,  Fig.  137,  be  parallel  to  V.  Then 
VQ  is  parallel  toi"  (§  102),  and  passes  through  the  F-trace  of 
the  line  B.  The  trace  HQ  is  determined,  as  in  the  general 
case,  by  the  if-traces  of  the  given  lines.  In  Eig.  137,  the 
point  in  which  HQ  and  VQ  intersect  on  GL  is  not  available, 
but  it  is  not  necessary. 

Special  Case  II.  If  one  of  the  given  lines,  as  A,  Fig.  138, 
is  parallel  to  both  H  and  V,  that  is,  parallel  to  the  ground  line, 


hq 


Ha 


Fia.  138. 


Ah 

a. 

r 

uf  "N 

Bh 

"*         \       \ 

A" 

l         '      / 
1         i  / 

ur '     J/. 

B" 

a. 
> 

i     7U1 

VQ 

A 
Fig. 

139. 

both  HQ  and  VQ  are  parallel  to  the  ground  line  (§  102),  one 
point  on  each  trace  being  located  by  the  line  B. 

Special  Case  III.  Suppose  both  given  lines  parallel  to  the 
ground  line.  Let  A  and  B,  Fig.  139,  be  the  given  lines;  then 
HQ  and  VQ  are  both  parallel  to  GL.  To  find  a  poinl  on  each, 
we  may  find  first  the  profile  trace  of  the  plane  Q  on  any 
assumed  profile  plane;  this  trace,  PQ,  will  pass  through  the 
profile  traces  up  and  v./,  of  A  and  B  respectively. 


82 


DESCRIPTIVE  GEOMETRY 


[X,  §  106 


Corollary  I.  To  find  the  plane  which  contains  a  given  line 
and  a  given  point. 

Analysis.  Through  the  given  point,  draw  an  auxiliary  line 
parallel  to  the  given  line,  thus  reducing  the  problem  to  one  of 
two  parallel  lines.  Or,  assume  any  point  on  the  given  line, 
and  draw  an  auxiliary  line  through  this  point  and  the  given 
point,  thus  reducing  the  problem  to  two  intersecting  lines.  The 
first  method  is  the  one  usually  adopted. 

Construction  (Fig.  140).  A  is  the  given  line,  c  the  given 
point.     The  auxiliary  line  B  is  drawn  through  c  parallel  to  A 


Fig.  140. 


(§  93),  and  the  required  plane  Q  is  passed  through  the  lines 
A  and  B. 

In  Fig.  141  the  line  A  is  parallel  to  V\  hence  the  direction 
of  VQ  is  known  to  be  parallel  to  Av  (§  102).  But  as  neither 
A  nor  B  has  a  F-trace,  a  point  on  VQ  is  determined  by  noting 
where  HQ  intersects  GL. 

Corollary  II.  To  find  the  plane  which  contains  three  given 
points  not  in  the  same  straight  line. 

Analysis.  Connect  any  two  points  by  a  straight  line,  and 
through  the  third  point  draw  a  second  line  parallel  to  the  first, 


X,  §  107]        PARALLEL  LINES  AND  PLANES 


83 


thus  reducing  the  problem  to  two  parallel  lines.  Or,  draw 
lines  connecting  any  point  with  each  of  the  other  two,  thus 
reducing  the  problem  to  one  of  two  intersecting  lines. 


^k\ 

\s>^ 

Cjf 

/ 1 
Xi 

/ 

/        4     / 

7 

V     ^> 

bv; 

Fig.  141. 


Fig.  142. 


No  figure  for  the  general  case  is  thought  necessary.  It  is 
interesting  to  note  that  this  device  may  be  used  for  passing  a 
plane  through  a  profile  line  and  a  point.  Thus  (Fig.  142),  let 
ab  be  the  given  line  and  c  the  given  point.  Draw  the  lines  ac 
(aW,  avcv),  and  be  (bhch,  bvcv),  and  pass  the  required  plane  Q 
through  these  lines.  This  avoids  the  necessity  of  finding  the 
traces  of  the  profile  line  ab,  although  its  traces,  if  found,  should 
lie  on  HQ  and  VQ  respectively  (§  96). 

107.  A  Plane  Parallel  to  Lines  or  to  Another  Plane.  A  plane 
which  is  parallel  to  given  lines  or  to  another  plane  is  deter- 
mined by  the  fact  that  a  plane  containing  one  of  two  parallel 
lines  is  parallel  to  the  other  line  (§  105). 

One  plane,  and  one  only,  may  be  found  which 

(a)  contains  a  given  line  and  is  parallel  to  a  second  given 
line  that  is  not  parallel  to  the  first  line ; 

(b)  contains  a  given  point  and  is  parallel  to  each  of  two  given 
lines  that  are  not  parallel  to  each  other ; 

(c)  contains  a  given  point  and  is  parallel  to  a  given  plane. 


84 


DESCRIPTIVE   GEOMETRY 


[X,  §  107 


Problem  7.  To  find  the  plane  which  contains  a  given  line  and  is 
parallel  to  a  second  given  line. 

Analysis.  Through  any  point  on  the  first  given  line  draw  an 
auxiliary  line  parallel  to  the  second  given  line.  The  required 
plane  is  determined  by  the  first  line  and  the  auxiliary  line. 


Fig.  143. 


Construction  (Fig.  143).  Let  it  be  required  to  pass  the  plane 
through  the  line  A.  Assume  any  point  con  J;  through  c  draw 
the  line  D  parallel  to  B.  Pass  the  required  plane  Q  through 
the  lines  A  and  D  (Prob.  6,  §  106). 


Fig.  144. 


X,  §  107]        PARALLEL  LINES  AND  PLANES 


85 


Special  Case  I.  Suppose  the  second  line  is  parallel  to  II 
or  V;  then  no  auxiliary  line  is  needed.  Thus,  in  Fig.  144,  let 
us  find  the  plane  which  contains  the  line  A  and  is  parallel  to 
B.  The  .ff-trace  of  any  plane  which  is  parallel  to  B  must  be 
parallel  to  Bh  (§  104).  Hence,  find  the  traces  of  A.  Draw 
HQ  through  sx  parallel  to  Bh ;  draw  VQ  through  tx  and  the 
point  in  which  HQ  intersects  GL. 

Special  Case  II.  Suppose  that  either  the  first  or  the  second 
given  line  is  a  profile  line.  The  general  solution  will  apply  to 
this  case  ;  but  if  the  problem  be  solved  in  this  manner,  a  profile 
projection  will  be  necessary.  A  simple  construction,  by  which 
the  use  of  a  profile  projection  may  be  avoided,  is  as  follows  : 

Let  ab  and  C,  Fig.  145,  be  the  given  lines.  Through  one 
point  of  the  profile  line,  as  a,  draw  an  auxiliary  line  D  parallel 
to  the  line  C;  through  the  other  point,  b,  of  ab,  draw  a  second 


Fig.  145. 


auxiliary  line  E  parallel  to  C.  Pass  the  plane  Q  through  the 
two  parallel  lines  D  and  E.  Then  if  the  required  plane  is  to 
contain  ab  and  be  parallel  to  C,  plane  Q  is  the  required  plane. 
If,  however,  the  required  plane  is  to  contain  C  and  be  parallel 
to  ab,  the  plane  It,  passed  through  C  parallel  to  plane  Q  (§  103), 
is  the  required  plane. 


86 


DESCRIPTIVE  GEOMETRY 


[X,  §  107 


Problem  8.  To  find  the  plane  which  contains  a  given  point  and  is 
parallel  to  each  of  two  given  lines. 

Analysis.  Through  the  given  point  draw  two  auxiliary  lines, 
one  parallel  to  one  given  line,  the  other  parallel  to  the  other 
given  line.  The  required  plane  is  determined  by  the  two  aux- 
iliary lines. 


Fig.  14(5. 


Construction  (Fig.  146).  Let  c  be  the  given  point,  A  and  B 
the  given  lines.  Through  c  draw  the  auxiliary  lines,  D  parallel 
to  A,  and  E  parallel  to  B.  Pass  the  required  plane  Q  through 
the  lines  D  and  E  (Prob.  6,  §  106). 

The  special  cases  of  this  problem  are  too  similar  to  those  of 
Problem  7  to  require  a  detailed  discussion. 

Problem  9.  To  find  the  plane  which  contains  a  given  point  and  is 
parallel  to  a  given  plane. 

Analysis.  Through  the  given  point,  pass  a  line  parallel  to 
the  given  plane.  Through  this  line,  pass  the  required  plane 
parallel  to  the  given  plane. 

Construction  (Fig.  147).  Let  Q  be  the  given  plane  and  a  the 
given  point.  A  line  may  be  drawn  through  a  and  parallel  to 
Q  by  drawing  it  parallel  to  either  the  horizontal  or  vertical 
principal  lines  of  Q  (§  104).  The  line  M,  drawn  through  a,  is 
parallel  to  the  vertical  principal  lines  of  Q ;  through  M  pass 
the  required  plane  T,  parallel  to  Q  (§  103). 


X,  §  107]       PARALLEL  LIXES  AND  PLANES 


87 


In  Fig.  148,  the  required  plane  T  is  found  by  means  of  the 
auxiliary  line  N,  which  is  parallel  to  the  horizontal  principal 
lines  of  the  given  plane  Q. 


Fig.  147. 


Fig.  148. 


Special  Case"  (Fig.  149).  Let  the  given  plane  Q  be  parallel 
to  the  ground  line.  Then  the  auxiliary  line  M,  drawn  through 
a  and  parallel  to  the  principal  lines  of  Q,  is  parallel  to  GL 
(§  101)  and  has  no  traces  on  H  and  V.     Pass  the  auxiliary 


Fki.  149. 


profile  plane  P  through  the  given  point.  Find  the  profile  trace, 
PQ,  of  Q,  and  the  profile  projection,  ap,  of  a.  The  profile 
trace,  PT,  of  the  required  plane  will  pass  through  a"  and  be 
parallel  to  PQ  (§  103).     From  PT  may  be  found  IIT  and  VT. 


88 


DESCRIPTIVE  GEOMETRY 


[X,  §  108 


108.    Use  of  Auxiliary  Lines  in  Finding  the  Traces  of  Planes. 

A  straight  line  is  determined  : 

1.  When  two  of  its  points  are  known ; 

2.  When  one  of  its  points  and  its  direction  (such  as  parallel 
or  perpendicular  to  another  line)  are  known. 

Eor  accuracy  of  construction,  the  second  method  of  deter- 
mining a  line  is  often  better  than  the  first.  In  any  event,  how- 
ever, a  line  is  not  determined  until  at  least  one  point  is  known. 

The  traces  of  the  required  planes  in  the  preceding  problems, 
being  straight  lines,  have  necessarily  been  located  by  one  of 
these  methods.  It  occasionally  happens,  however,  that  neces- 
sary points  for  locating  these  traces  fall  far  outside  any  rea- 
sonable limits  for  the  size  of  the  figure,  and  recourse  must  be 
had  to  auxiliary  lines  which  will  locate  points  within  reach. 

Example  1  (Eig.  150).  Let  it  be  required  to  find  the  plane 
which  contains  the  lines  A  and  B,  so  situated  that  only  one 


Fig.  150. 


point,  s2,  on  HQ,  and  one  point,  t2,  on  VQ,  can  be  found. 
Assume  any  point,  e,  in  the  given  line  A ;  through  this  point 
draw  the  line  D,  parallel  to  the  given  line  B.  Pass  the  plane 
Q  through  the  parallel  lines  B  and  D.  This  plane  must 
necessarily  contain  the  line  A,  since  A  intersects  both  B  and 
D,  and  is  the  required  plane. 

In   general,  assume  any  point   on  one   of   the  given  lines. 


X,  §  108]        PARALLEL  LINES  AND  PLANES 


89 


Through  this  point  draw  an  auxiliary  line  parallel  to  the 
second  given  line ;  the  auxiliary  line  will  lie  in  the  plane  of 
the  given  lines.  Judgment  must  be  used  in  selecting  the  line 
to  which  the  auxiliary  line  should  be  made  parallel,  in  order  that 
this  latter  line  may  be  of  service  in  obtaining  additional  points. 
Example  2.  Another  method  of  obtaining  a  line  lying  in 
the  plane  of  two  given  lines  is  shown  in  Fig.  151.  Let  A  and 
B  be  the  given  lines.  Assume  any  point,  e,  in  A,  and  any 
point,  /,  in  B.  Then  the  line  D,  passing  through  e  and  /,  will 
lie  in  the  plane  of  the  lines  A  and  B,  and  the  traces  of  D  will 
lie  in  the  traces  of  the  required  plane  Q.  As  before,  judg- 
ment must  be  used  in  selecting  the  points  e  and  /. 


Fig.  151. 


Fig.  152. 


Example  3.  When  the  given  data  are  sufficient  to  determine 
one  trace  of  the  plane,  a  very  useful  auxiliary  line  is  one  of 
the  principal  lines  (§  99)  of  the  plane  itself.  Let  A  and  B, 
Fig.  152,  be  sufficient  to  locate  VQ  but  not  HQ.  Assume  a 
point  in  one  of  the  given  lines,  as  point  c  in  line  A.  Through 
c  draw  a  vertical  principal  line  of  Q,  by  making  Mv  parallel  to 
VQ,  and  M h  parallel  to  GL.  Then  the  line  M  lies  in  Q,  and 
its  H-tra.ce,  s3,  is  a  point  in  HQ. 

If  HQ  were  the  known  trace,  a  horizontal  principal  line  of 
the  plane  could  be  similarly  drawn  to  locate  a  point  in  VQ' 


90 


DESCRIPTIVE  GEOMETRY 


[X,  §  108 


Example  4.  Another  method  of  procedure,  when  one  trace 
of  the  plane  is  known,  is  shown  in  Fig.  153.  Let  VQ  be  the 
known  trace.  Assume  any  point,  tv,  in  VQ  ;  this  point  lies  in 
V,  and  th  is  in  GL.  Through  the  point  t,  draw  the  auxiliary 
line  D  parallel  to  one  of  the  given  lines,  as  A.  Then  the  line 
D  lies  in  the  plane  Q,  and  its  //-trace,  s*,  is  a  point  in  HQ. 


Fig.  153. 


Fig.  154. 


In  general,  the  auxiliary  line  D  may  be  drawn  parallel  to 
either  of  the  given  lines  A  or  B ;  but  in  this  case  D  should 
obviously  be  drawn  parallel  to  A  in  order  to  obtain  a  result 
within  the  limits  of  the  figure. 

Example  5.  Let  the  lines  A  and  B,  Fig.  154,  be  sufficient 
to  locate  HQ  but  not  VQ.  A  third  method  of  obtaining  an 
auxiliary  line  is  to  assume  any  point,  as  s,  on  HQ,  and  any 
point,  as  e,  on  either  of  the  given  lines  A  or  B.  Then  the  line 
I),  joining  s  and  e,  lies  in  the  plane  Q,  and  its  F-trace,  tv,  is  a 
point  in  VQ. 

In  some  situations,  in  order  to  get  a  sufficient  number  of 
points,  it  may  be  necessary  to  employ  more  than  one  auxiliary 
line.  This  would  have  been  the  case  in  Figs.  152,  153,  and 
154,  if  in  any  one  of  these  figures  the  intersection  of  HQ  and 
VQ  with  the  ground  line  had  not  been  available. 


CHAPTER   XI 


PERPENDICULAR    LINES    AND    PLANES 


109.  Perpendicular  Lines.  If  two  straight  lines  are  perpen- 
dicular to  each  other,  the  fact  is  not,  in  general,  apparent  from 
the  projections  of  the  lines.  For  example,  consider  the  square 
prism  represented  in  Fig.  94.     The  edges  of  this  solid  form 


Fig.  M  (repeated). 

three  series  of  mutually  perpendicular  lines;  but  all  these 
edges  are  oblique  to  V,  and  in  the  ^projection  no  right  angles 
appear. 

In  the  II-  and  secondary  ^projections,  however,  the  pro- 
jected edges  are  at  right  angles.  This  shows  that,  under  cer- 
tain conditions,  it  is  possible  to  recognize  perpendicular  lines 
from  their  projections. 

91 


92 


DESCRIPTIVE  GEOMETRY 


[XI,  §  110 


110.  Perpendicular  Lines  Whose  Projections  Are  Perpendicular. 
If  two  lines  in  space  are  mutually  perpendicular,  and  are  pro- 
jected on  any  plane  of  projection  parallel  to  one  of  the  lines,  the 
two  projections  will  he  perpendicular  to  each  other. 

There  is  one  and  only  one  exception.  The  plane  of  projec- 
tion must  not  be  perpendicular  to  the  second  line  ;  for  in  this 
case  the  line  would  project  as  a  point,  and  a  point  cannot  be 
said  to  be  perpendicular  to  a  line. 

Let  the  line  A,  Fig.  155,  be  parallel  to  H.  Let  Q  be  a  plane 
perpendicular  to  A  and  consequently  perpendicular  to  H.  In 
this  position  of  the  line  and  plane,  it  is  evident  that  Ah  and 
HQ  are  perpendicular  to  each  other.  Now  let  B  be  any  line 
drawn  in  the  plane  Q.  Since  A  is  perpendicular  to  Q,  the 
lines  A  and  B  must  be  mutually  perpendicular.     Also,  since 


Fig.  155. 


Fig.  15li. 


Q  is  perpendicular  to  H,  Bh  must  coincide  with  II Q ;  that  is, 
the  projections  Ah  and  Bh  are  perpendicular  to  each  other.  It 
may  be  noted  that  the  line  B  need  not  necessarily  intersect  the 
line  A,  since  any  line  lying  in  Q  is  considered  to  be  perpen- 
dicular to  A. 

What  is  true  of  the  if-plane  of  projection  must  be  true  of 
any  plane  of  projection,  hence  the  proposition  is  established. 

111.  A  Line  Perpendicular  to  a  Plane.  If  a  line  is  perpen- 
dicular to  a  plane,  any  projection  of  the  line  is  perpendicular  to 
the  corresponding  trace  of  the  plane. 

Let  the  line  A,  Fig.  156,  be  perpendicular  to  the  plane  Q. 
Then  A  will  be  perpendicular  to  any  line  lying  in  Q,  in  par- 
ticular to  the  horizontal  principal  line  B.     But  B  is  parallel  to 


XI,  §  113]     PERPENDICULAR  LINES  AND  PLANES      93 

H;  hence  Ah  is  perpendicular  to  Bh,  by  §  110.  Moreover  Bh  is 
parallel  to  HQ.  Hence  Ah  is  perpendicular  to  HQ.  Similarly 
for  any  other  plane  of  projection. 

112.  Test  for  Perpendicularity  of  a  Line  and  a  Plane.  Con- 
versely, if  the  horizontal  and  vertical  projections  of  a  line  are 
perpendicular  to  the  corresponding  traces  of  a  plane,  the  line  and 
plane  are  mutually  perpendicular. 

The  only  exception  occurs  when  the  plane  is  parallel  to  the 
ground  line,  for  the  projections  of  any  profile  line  on  H  and  V 
are  perpendicular  to  the  H-  and 
F-traces  of  such  a  plane.  Such 
cases  may  be  tested  by  the  use 
of  the  profile  projection.  Thus,  in 
Fig.  157,  the  line  ab  is  found  to 
be  perpendicular  to  the  plane  Q. 

113.  Perpendicular  Planes.  If 
two  planes  are  mutually  perpen- 
dicular, the  fact  is  not,  in  general, 
evident  from  the  traces  of  the 
planes.  Thus,  in  Fig.  158,  the 
line  A  is  perpendicular  to  the  plane  Q.  Hence  any  plane,  as 
R  or  S,  passed  through  A,  is  perpendicular  to  Q ;  but  this 
relation  cannot  be  seen  directly  from  the  traces  of  the  planes. 


Fig.  158. 


Fig.  15it. 


But  let  the  line  A,  Fig.  159,  be  perpendicular  to  the  plane 
Q,  and  let  us  pass  through  A  the  plane  T  perpendicular  to  //. 


94  DESCRIPTIVE   GEOMETRY  [XI,  §  113 

Since  HT  coincides  with  Ah,  and  Ah  is  perpendicular  to  HQ, 
it  follows  that  HT  is  perpendicular  to  HQ.  Hence  we  have 
the  proposition  : 

If  two  planes  are  mutually  perpendicular,  and  one  of  them  is 
perpendicular  to  a  coordinate  plane,  the  traces  of  the  two  given 
planes  on  that  coordinate  plane  are  perpendicular. 

114.  Lines  of  Maximum  Inclination  to  H  and  V.  Let  Q  be 
any  plane  oblique  to  H.  Those  of  its  lines  which  are  perpen- 
dicular to  HQ  have  the  peculiarity  that  they  make  with  H  a 
greater  (acute)  angle  than  any  other  lines  in  the  plane.     For 

this  reason  they  are  called  the  lines 

X.  of  maximum  inclination  to  H  of  the 

\  plane  Q.     Let  M,  Fig.  160,  be  one 

V \ob  °f  these  lines.     Since  M  is  perpen- 

V/\  dicular  to  HQ,  it  follows  that  Mh 

\       \^  must  also  be  perpendicular  to  HQ. 

N.  ^      /  A  plane  is  determined  when  one 

>^  J  /  of  its  lines  M  (Mh,  M")  of  maximum 

y^  inclination  to  H  is  given.     For  sup- 

^y  pose  that  s  and  t,  Fig.  160,  are  the 

„      ,_.  traces    of    this    line.       Then    the 

Fig.  160. 

//-trace,  HQ,  of  the  required  plane 
passes  through  s  and  is  perpendicular  to  Mh,  while  the  F-trace, 
VQ,  is  determined  by  t  and  the  point  in  which  HQ  cuts  the 
ground  line. 

A  similar  analysis  applies  to  lines  of  maximum  inclination 
to  F 

115.  Planes   Perpendicular  to  a  Given  Line  or  Plane.      One 

plane,  and  one  only,  may  be  found  which  contains  a  given  point 
and  is  perpendicular  to  a  given  straight  line.  The  point  may 
be  on  the  line,  or  at  any  distance  from  it. 

One  plane,  and  one  only,  may  be  found  which  contains  a 
given  straight  line  and  is  perpendicular  to  a  given  plane.  An 
exception  occurs  when  the  line  is  itself  perpendicular  to  the 
given  plane,  in  which  case  an  infinite  number  of  planes  may  be 
found. 


XI,  §  115]     PERPENDICULAR  LINES  AND  PLANES       95 


Problem  10.  To  find  the  plane  which  contains  a  given  point  and 
is  perpendicular  to  a  given  line. 

Analysis.  Through  the  given  point  draw  an  auxiliary  line 
which  shall  be  perpendicular  to  (but  does  not  necessarily  inter- 
sect) the  given  line.  Through  this  line  pass  the  required  plane 
perpendicular  to  the  given  line. 

Construction  (Fig.  161).  Let  c  be  the  given  point  and  A  the 
given  line.  Through  c  draw  the  auxiliary  line  M,  making  M h 
perpendicular  to  Ah  and  Mv  parallel  to  OL.  Then  the  line  M 
is  parallel  to  H,  and  consequently  perpendicular  to  A  (§  110). 
Find  the  F-trace,  t,  of  M.     Through  t  draw  VQ  perpendicular 


b- 

\na 

CLh 

a.      N       % 
-1            S^        > 

Fig.  161. 


Fig.  102. 


to  Av,  and  from  the  point  in  which  VQ  intersects  GL  draw 
HQ  perpendicular  to  Ah.  Then  Q  is  perpendicular  to  A 
(§  112)  and  is  the  required  plane. 

Note  that  the  auxiliary  line  M  is  a  horizontal  principal  line 
of  the  required  plane  Q.  The  construction  could  also  be 
effected  by  using  an  auxiliary  line  parallel  to  V;  such  a  line 
would  be  a  vertical  principal  line  of  Q. 

Special  Case.  The  general  method  fails  when  the  given 
line  is  a  profile  line.  Let  ab,  Fig.  102,  be  the  given  line,  and 
c  the  given  point.  The  required  plane  Q  will  be  parallel  to 
the  ground  line,  hence  perpendicular  to  P.  Find  the  profile 
projection  of  the  line  ab  and  the  point  c.  Through  c"  draw 
the  edge  view  and  profile  trace,  PQ,  of  Q,  perpendicular  to 
apbp.     From  PQ  find  J/Q  and  VQ. 


96 


DESCRIPTIVE   GEOMETRY 


[XI,  §  115 


Problem  11.  To  find  the  plane  which  contains  a  given  line  and  is 
perpendicular  to  a  given  plane. 

Analysis.  Through  any  point  of  the  given  line  draw  an 
auxiliary  line  perpendicular  to  the  given  plane.  Any  plane 
containing  this  line  will  be  perpendicular  to  the  given  plane 
(§  113).  Hence  the  required  plane  is  the  plane  determined 
by  the  given  and  auxiliary  lines. 

Construction  (Fig.  163).  Let  A  be  the  given  line  and  Q  the 
given  plane.     Assume  any  point  c  on  the  line  A,  and  through 


Fig.  163. 


Fig.  Ifi4. 


c  draw  the  auxiliary  line  B  perpendicular  to  Q  (§  112).  Find 
the  plane  T,  containing  the  lines  A  and  B  (Prob.  6,  §  106) ; 
this  is  the  required  plane. 

Special  Case  I.  If  the  given  line  is  a  profile  line,  the  gen- 
eral method  applies  ;  but  the  use  of  a  profile  projection  may  be 
avoided  as  follows.  Let  ab,  Fig.  164,  be  the  given  line,  and  Q 
the  given  plane.  From  each  end  of  the  given  line  draw  the 
respective  auxiliary  lines  D  and  E,  each  perpendicular  to  Q. 
Pass  the  required  plane  T  through  the  parallel  lines  D  and  E 
(compare  Fig.  145). 

Special  Case  II  (Fig.  165).  Let  the  given  plane,  Q,  be 
parallel  to  the  ground  line,  and  let  A  be  the  given  line.     As  in 


XI,  §  115]     PERPENDICULAR  LINES  AND  PLANES       97 


the  general  case,  assume  any  point  c  in  the  line  A,  and  draw 
the  line  B  perpendicular  to  Q.  Then  the  required  plane,  T, 
is  the  plane  determined  by  the  lines  A  and  B.  To  find  the 
traces  of  T  we  need  the  traces  of  these  two  lines.  The  traces 
of  A  are  readily  obtained  (Prob.  1,  §  37).     To  find  the  traces 


0- 
H*     X 

Id 
V  / 

^C  /\:  vol 

¥    x 

c 

0. 
> 

r                V 

Fig.  105. 

of  By  we  note  that  B  is  a  profile  line  and  that  we  know 
only  one  point,  c,  in  this  line.  Nevertheless,  B  is  a  definite 
line  in  space,  since  its  direction,  perpendicular  to  the  plane  Q, 
is  known.  Hence,  find  the  profile  projection,  cp,  of  the  point 
c,  and  the  profile  trace,  PQ,  of  the  plane  Q.  Then  the  profile 
projection,  Bp,  of  line  B  is  drawn  through  cp  perpendicular  to 
PQ.  From  B"  we  can  find  the  //-  and  F-traces  of  line  B. 
These  traces,  together  with  the  traces  of  line  A,  determine  the 
traces  of  the  required  plane  T. 


CHAPTER   XII 

INTERSECTION   OF  PLANES   AND  OF   LINES   AND  PLANES 
—  APPLICATIONS 

116.  Intersecting  Planes.  Let  two  planes  Q(HQ,  VQ)  and 
R  (HR,  VR)  intersect  in  a  line  A.  The  figure  will  be  left  for 
the  student  to  draw.  One  point  of  the  intersection  will  be 
determined  if  we  find  where  a  line  in  one  plane  intersects  a 
line  in  the  other.  A  second  point  may  be  similarly  found  by 
the  intersection  of  a  second  pair  of  lines.  Evidently  any  such 
pair  of  lines  cannot  be  chosen  at  random,  for  in  that  case  they 
probably  would  not  intersect.  The  //-traces  of  the  two  planes, 
however,  as  they  are  both  in  H,  will  in  general  intersect. 
Likewise  the  two  F"-traces  will  in  general  intersect.  But 
the  intersection  of  the  i7-traces  will  be  the  //"-trace  of  the 
required  line  .1  (§96),  and  the  intersection  of  the  p^traces 
will  be  the  F-trace  of  A.  The  projections  of  the  line  A  may 
then  be  determined  (Prob.  2,  §  37). 

117.  A  Point  in  the  Intersection  of  Two  Planes.  The  general 
method  of  obtaining  a  point  in  the  line  of  intersection  of  two 
planes  is  as  follows  :  Let  two  planes  Q  and  R,  as  before,  inter- 
sect in  a  line  A,  and  let  X  .be  any  plane  not  parallel  to  A. 
Then  X  will  intersect  Q  in  a  line  xq,  and  will  intersect  R  in  a 
line  xr.  Since  these  lines  are  in  the  same  plane,  X,  and  can- 
not be  parallel  (why  ?),  they  intersect  in  a  point,  xqr,  which 
lies  in  all  three  planes,  A",  Q,  and  R.  Hence  xqr  is  a  point  in 
the  line  of  intersection,  A,  of  Q  and  R. 

118.  The  Line  of  Intersection  of  Two  Planes.  The  line  of 
intersection  of  two  planes  becomes  known:  (1)  when  two 
points  of  the  line  are  known ;  (2)  when  one  point  and  the 
direction  of  the  line  are  known.     (Compare  §  108.) 

Problem  12.     To  find  the  line  of  intersection  of  two  planes. 
Analysis.     The  traces    of  the  required  line  of   intersection 
lie  at  the  intersection  of  the  corresponding  traces  of  the  given 

98 


XII,  §  118]        INTERSECTION  OF  PLANES 


99 


planes  (§  116).     The  line  may  now   be  determined   from  its 
traces  (Prob.  2,  §  37). 

Construction,  General  Case  (Fig.  166).  Let  Q  and  R  be  the 
given  planes.  The  construction  should  be  evident  from  the 
analysis. 


Fig.  1G0. 


Fig.  167. 


A  second  example,  with  planes  differently  situated,  is  given 
in  Fig.  167. 

The  general  case  may  fail  because  of  (a)  parallel  lines ; 
(6)  intersections  inaccessible  ;  (c)  points  s  and  t  coincident. 

Special  Case  I.  Suppose  that  one  pair  of  traces  is  parallel 
(Fig.  168).  Let  Q  and  R  be  the  given  planes,  with  HQ  and  11 R 
parallel.  The  intersection  of  VQ 
and  VR  gives  the  F-trace,  t,  of  the 
required  line  of  intersection  A. 
Consider  the  planes  Q,  R,  and  //. 
Since  the  intersections  of  H  with 
Q  and  R  are  parallel,  H  must  be 
parallel  to  the  line  of  intersection, 
A,  of  these  planes.  Hence  Ah 
passes  through  th  and  parallel  to 
HQ  and  II R,  while  A"  passes  through  t"  and  is  parallel  to  GL. 
The  line  of  intersection  is  thus  seen  to  be  a  horizontal  principal 
line  (§  99)  of  each  of  the  given  planes. 

If  the  F-traces  of  the  given  planes  were  parallel,  while  the 
//-traces  intersected,  the  resulting  line  of  intersection  would 
be  parallel  to  V,  a  vertical  principal  line  of  each  of  the  gives 
planes. 


Fig.  Ifi8. 


100 


DESCRIPTIVE    GEOMETRY        [XII,  §  118 


Q. 

I 

Ha 

VR    \ 

Ah       uh 

\                           \ 

If  both  the  H-  and  F"-traces  of  the  given  planes  are  respec- 
tively parallel,  the  planes  are  parallel  (§  103),  except  in  the 
case  in  which  each  plane  is  parallel  to  the  ground  line. 

Special  Case  II.  Suppose  that  both  planes  are  parallel  to 
the  ground  line  (Fig.  169).     Let  Q  and  R  be  the  given  planes. 

One  point  of  the  line  of  intersec- 
tion may  be  obtained  by  finding 
its  profile  trace  on  any  profile 
plane  of  projection.  Let  P(HP, 
VP)  be  any  assumed  profile 
plane.  Find  the  profile  traces 
PQ  and  PR  of  the  given  planes 
(§  60).  Their  intersection  gives 
up,  the  profile  trace  of  the  re- 
quired line  of  intersection  A. 
From  u"  obtain  uh  and  uv,  one 
point  in  the  line  A.  A  second 
point  is  not  necessary,  since  this 
line  must  be  parallel  to  both  H  and  V  (Case  I),  that  is,  parallel 
to  the  ground  line. 

Special  Case  III.  One  plane  is  parallel  to  H  or  V,  the 
other  is  oblique  (Fig.  170).     Let  Q  and  R  be  the  given  planes, 


\N        HR       > 

\    v                   ' 

Av 

y\9 

/      \      VQ. 

a. 
> 

Fig.  109. 


Fig.  170 


the  plane  R  being  parallel  to  H.     The  plane  R  therefore  inter- 
sects Q  in  a  horizontal  principal  line  of  Q  (§  99). 

If  the  plane  R  be  taken  parallel  to  V,  the  intersection  be- 
comes a  vertical  principal  line  of  Q. 


XII,  §  118]        INTERSECTION  OF  PLANES 


101 


This  is  an  important  case,  since,  on  account  of  the  simplicity 
of  the  intersection,  planes  parallel  to  H  or  V  are  often  intro- 
duced as  auxiliaries  when  the  general  case  fails. 

Special  Case  IV.  The  traces  are  not  parallel,  but  one  or 
both  pairs  fail  to  intersect  within  the  limits  of  the  drawing. 
Let  Q  and  R,  Fig.  171,  be  the  given  planes.  The  intersection 
of  the  P~-traces  of  these  planes  gives  the  point  t,  as  in  the 
general  case.  Since  HQ  and  HR  do  not  intersect  within 
reach,  pass  the  auxiliary  plane,  X,  parallel  to  H.  Planes  X 
and  Q  intersect  in  the  line  M  (Case  III) ;  planes  X  and  R 
intersect  in  the  line  X;  the  lines  M  and  N  intersect  in  the 
point  c,  which  is  a  point  in  the  intersection  of  the  planes  Q 
and  R  (§  117).  The  points  t  and  c  determine  the  required  line 
of  intersection  A. 

In  Fig.  172,  neither  pair  of  traces  intersect  within  the  limits 
of  the  figure.  Two  auxiliary  planes,  X  and  Y,  are  used,  each 
of  which  locates  a  point  on  the  line  of  intersection  A.  The 
auxiliary  planes  may  be  parallel  to  either  H  or  V;  in  the 
figure  they  are  taken  parallel  to  V. 


Fig.  172. 


Fig.  17:5. 


Special  Case  V.  Both  planes  intersect  the  ground  line  at 
the  same  point  (Tig.  173).  Let  Q  and  R  be  the  given  planes. 
The  general  r-ase  fails  because  the  points  sand  t  are  coincident. 
A  point,  c,  in  the  required  line  of  intersection,  A,  may  be 
determined  by  means  of  the  auxiliary  plane  X  parallel  to  V,  as 
in  Fig.  172.  An  auxiliary  plane  parallel  to  H  may  be  used  if 
prefer  icd. 


102 


DESCRIPTIVE  GEOMETRY 


[XII,  §  118 


Special  Case  VI.  One  of  the  planes  contains  the  ground 
line  (Fig.  174).  Let  Q  and  R  be  the  given  planes.  R  passes 
through  the  ground  line,  and  is  determined  by  the  quadrants 

through  which  it  passes 
and  its  angle  with  H.  As 
in  Special  Case  V,  the 
points  s  and  t  are  coinci- 
dent on  the  ground  line. 
An  additional  point  in  the 
required  line  of  intersec- 
tion may  be  found  by  the 
use  of  an  auxiliary  profile 
plane  of  projection,  as- 
sumed anywhere  except 
through  the  coincident 
points  s  and  t.  Find  the 
profile  traces,  PQ  and  PR, 
of  the  given  planes  (§  60).  These  traces  intersect  in  cp, 
which  is  the  profile  projection  of  a  point  in  the  required  line 
of  intersection. 

119.  The  Intersection  of  a  Line  and  a  Plane.  Let  a  line  A 
(Fig.  175)  intersect  a  plane  Q.     The  point  of  intersection  will 


R  is  in  quad- 
rants  I  %-  3,  and 
makes    60°  with  H 


Fig.  174. 


Fig.  175. 

be  determined  if  we  find  where  A  intersects  a  line  in  plane  Q. 
This  line  cannot,  however,  be  any  line  chosen  at  random  in  Q, 
for  such  a  line  will  probably  not  intersect  A.  Let  a  plane,  X, 
be  passed  through  the  line  A.  Then  X  will  intersect  Q  in  a 
line,  T.     This  line  Y  is  a  line  in  the  plane  Q,  which  is  inter- 


XII,  §  119]        INTERSECTION  OF  PLANES  103 

sected  by  A  at  the  point  c.  Hence  c  is  the  required  point  in 
which  A  intersects  Q.  The  solution  thus  depends  directly  upon 
the  preceding  problem. 

Problem  13.  To  find  the  point  in  which  a  straight  line  intersects  a 
plane. 

Analysis.  Through  the  given  line,  pass  any  auxiliary  plane. 
Find  the  line  of  intersection  of  the  auxiliary  plane  with  the 
given  plane,  and  note  the  point  in  which  this  line  intersects 
the  given  line.  This  is  the  required  point  of  intersection  of 
the  given  line  and  plane. 

General  Method.  Construction  (Fig.  176).  Let  A  be  the  given 
line,  and  Q  the  given  plane.     Through  A  pass  any  auxiliary 


Fig.  176. 

plane,  as  X  (§  98).  Find  the  line  of  intersection,  B,  of  X  and 
Q  (Prob.  12,  §  118).  The  line  A  intersects  the  line  B  in  the 
point  c,  the  required  point  of  intersection  of  A  and  Q. 

Check.  The  projection  ch  is  determined  by  the  intersection 
of  Ah  and  Bh;  C  is  independently  determined  by  the  intersec- 
tion of  Av  and  B".  But  c*  and  c"  are  two  projections  of  the 
same  point,  and  hence  must  lie  in  the  same  projector. 

Since  the  auxiliary  plane  X  may  be  any  plane  passed 
through  the  line  A,  ease  of  construction  often  depends  upon  a 
judicious  choice  of  this  plane.  Ordinarily,  the  simplest  con- 
struction results  when  the  auxiliary  plane  is  taken  perpen- 
dicular to  H  or  V  (see  the  planes  X  and  Y,  Fig.  125). 


104 


DESCRIPTIVE  GEOMETRY 


[XII,  §  119 


Usual  Method.  Construction  (Fig.  177).  Let  A  be  the  given 
line  and  Q  the  given  plane.  The  auxiliary  plane  X  is  taken 
through  A  and  perpendicular  to  H.  Planes  X  and  Q  intersect 
in  the  line  B.  The  projections  Av  and  B"  intersect  in  cv,  one 
projection  of  the  required  point.  The  projections  Ah  and  Bh  are 
coincident,  so  that  their  intersection  is  indeterminate.  Conse- 
quently the  projection  ch  must  be  located  by  projecting  from  cv. 

The  auxiliary  plane  X  might  otherwise  have  been  taken  per- 
pendicular to  V,  in  which  case  ch  would  be  determined  by  the 
direct  intersection  of  two  projections,  while  c"  would  need  to 


Fig.  177. 


Fig.  178. 


be  located  by  projection  from  c\  The  usual  method  does  not 
give  the  check  on  the  construction  which  appears  in  the  general 
method.  In  applications,  however,  the  gain  in  simplicity 
usually  more  than  compensates  for  this. 

Special  Case.  The  given  plane  is  perpendicular  to  H  or  V. 
Let  the  given  plane  Q,  Fig.  178,  be  perpendicular  to  V.  ISTo 
construction  is  necessary,  since  VQ  is  an  edge  view  of  the 
plane,  and  in  the  ^projection  the  point  in  which  the  line  A 
pierces  the  plane  appears  directly.  The  //-projection  of  this 
point  is  obtained  by  projecting  from  the  F-projection. 

120.    The  Shortest  Distance  from  a  Point  to  a  Plane.     The 

shortest  distance  from  a  given  point  to  a  given  plane  may  be 
obtained  by  dropping  a  perpendicular  from  the  point  to  the 
plane,  and  then  measuring  the  length  of  this  perpendicular. 


XII,  §  120]        INTERSECTION    OF  PLANES 


105 


Problem  14.  To  find  the  shortest  distance  from  a  point  to 
a  plane. 

Analysis.  From  the  given  point  drop  a  perpendicular  to  the 
given  plane.  Find  the  foot  of  the  perpendicular,  that  is,  the 
point  in  which  the  line  pierces  the  given  plane.  Obtain  the 
true  length  of  the  perpendicular. 

Note.  Observe  tbat  this  solution  is  a  direct  application  of  the  previous 
Problem. 

Construction  (Fig.  179).  Let  a  be  the  given  point,  and  Q  the 
given  plane.  From  a  draw  the  indefinite  line,  O,  perpendicular 
to  Q  (§  111).     Find   the    point,    b,    in   which    C  intersects  Q 


Fig.  179. 


Fig.  180. 


(Prob.  13,  §  119) ;  in  the  figure,  this  is  done  by  using  the  aux- 
iliary plane  X,  perpendicular  to  II.  Then  ahbh  and  avbv  are  the 
projections  of  the  required  shortest  distance,  the  true  length 
of  which  may  be  found  by  Problem  3  (§  78  or  §  80). 

Special  Case  i  Fig.  180).  The  given  plane  Q  is  parallel  to 
the  ground  line.  The  required  perpendicular  from  a  is  evi- 
dently a  profile  line,  and  may  be  drawn  directly  in  the  profile 
projection  as  soon  as  the  profile  views  of  the  given  point  and 
and  plane  are  obtained.  In  (ids  case  it  is  not  necessary  to 
have  the  H-  and  ^projections  of  the  perpendicular  in  order  to 
know  its  actual  length  ;  these  projections  are,  however,  usually 
considered  a  part  of  the  problem,  and  are  obtained  bj  project- 
ing from  the  profile  view. 


106 


DESCRIPTIVE   GEOMETRY 


[XII,  §  121 


121.    The    Projection    of   a   Point   or   Line   on  a  Plane.     The 

projection  of  a  point  on  a  plane  is  the  foot  of  the  perpendicular 
dropped  from  the  point  to  the  plane.  This  definition  is  not 
confined  to  the  coordinate  planes  of  projection,  but  applies  to 
any  plane  in  space.  However,  when  a  point  is  projected  on  to 
some  oblique  plane  represented  by  its  traces  on  H  and  V,  the 
projection  must  in  turn  be  represented  by  its  projections  on  H 
and  V.  Thus,  in  Figs.  179  and  180,  the  point  b  (bh,  bT)  is  the 
projection  of  the  point  a  (ah,  av)  on  the  plane  Q  (HQ,  VQ). 

The  projections  of  a  straight  line  may  be  obtained  by  pro- 
jecting any  two  of  its  points,  and  then  connecting  these  pro- 
jections. 

Problem  15.     To  project  a  line  on  an  oblique  plane. 

Analysis.  Erom  any  two  points  on  the  given  line,  drop  per- 
pendiculars to  the  given  plane.     Eind   the   points   in   which 


Fig.  181. 


these  lines  intersect,  respectively,  the  given  plane,  and  connect 
these  points. 

Construction  (Fig.  181).  Let  it  be  required  to  project  the 
line  ab  on  the  plane  Q.  From  a  draw  the  line  C  (Ch,Cv)  per- 
pendicular to  the  plane  Q  (§  111),  and  find  the  point,  e,  where 
C  intersects  Q  (Prob.  13,  §  119).  From  b  draw  the  line  D 
perpendicular  to  Q,  and  find  point  /,  where  D  intersects 
Q.     Then  the  line  ef  (eHfh,  evfv)  is  the  projection  of  ab  on  Q. 

A  second  example  is  given  in  Fig.  182,  the  lettering  and 
explanation  being  the  same  as  for  the  preceding  figure.     The 


XII,  §  121]        INTERSECTION  OF  PLANES 


107 


fact  that  the  projection  ef  crosses  the  given  line  ab  shows  that 
this  line  intersects  the  plane  Q  at  the  point  n,  where  ab  and  ef 
intersect.     Note  that  the  projections  nh  and  nv  are  independ- 


Fig.  1S2. 


ently  determined,   and  should   check  by   lying  in    the  same 
projector. 

Special  Case.     In  Fig.  183  the  given  line  A  is  known  to  be 
parallel  to  the  given  plane  Q  (§  104).     Hence,  to  project  A  on 


Fm.  183. 


Q  it  is  sufficient  to  project  one  point  of  A,  as  for  example, 
point  c,  which  projects  at  point  c.  Then  the  required  line  B  is 
drawn  through  e  parallel  to  the  line  A. 


CHAPTER    XIII 

INTERSECTIONS   OF  PLANES   AND   SOLIDS   BOUNDED   BY 
PLANE   FACES 

122.  A  Plane  Determined  by  Two  Lines.  A  plane  is  com- 
pletely determined  when  any  two  of  its  lines,  not  necessarily 
its  traces,  are  known  (§  106).  Hence,  if  a  plane  be  given  by 
means  of  any  two  intersecting  or  parallel  lines  in  space,  it  is 
not  always  necessary,  nor  even  desirable,  to  find  its  traces  in 
the  solution  of  a  problem  in  which  such  a  plane  occurs. 

123.  The  Intersection  of  a  Line  with  a  Plane  Determined  by 
Two  Lines.     Let  it  be  recpiired  to  find  the  point  in  which  the 


Fig.  184. 

line  C,  Fig.  184,  intersects  the  plane  of  the  intersecting  lines 
A  and  B.  Instead  of  finding  the  traces  of  the  plane  containing 
A  and  B  (Prob.  6,  §  106),  and  then  finding  the  point  in  which  0 
intersects  this  plane  (Prob.  13,  §  119),  let  us  proceed  at  once,  as 
in  the  usual  method  of  Problem  13,  to  pass  through  the  line 

108 


XIII,  §  123]        INTERSECTION  OF  PLANES 


109 


C  an  auxiliary  plane  perpendicular  to  H  .(or  to  V).  The  line 
HX,  coincident  with  Ch,  is  the  TZ-trace  and  edge  view  of  such 
a  plane.  The  plane  X  intersects  the  line  A  in  point  j  (see 
Fig.  178)  and  the  line  B  in  point  k.  The  line  D,  connecting  j 
and  k,  must  therefore  be  the  line  of  intersection  of  the  plane 
X  with  the  plane  of  the  lines  A  and  B.  The  projection  Dv 
intersects  C"  at  ev,  which,  for  the  same  reasoning  as  that  given 
in  Problem  13,  must  be  one  projection  of  the  point  in  which 
C  intersects  the  plane  of  A  and  B.  Finally,  eh  is  found  by  pro- 
jecting from  ev. 

Note  that  in  this  solution  the  position  of  the  ground  line  is 
not  essential.     It  may  therefore  be  omitted. 

Figure  185  shows  how  to  apply  this  method  to  find  the 
point  in  which  the  profile  line  cd  pierces  the  plane  of  the  in- 


tersecting lines  A  and  B.  Pass  through  cd  a  profile  plane. 
The  plane  of  A  and  B  intersects  the  profile  plane  in  the 
line  ab.  By  means  of  a  profile  projection  it  is  readily  found 
that  the  lines  ab  and  cd  intersect  in  the  point  e,  the  required 
piercing  point.  In  this  figure  the  ground  line  is  not  omitted, 
since  it  is  necessary  in  order  to  find  the  profile  projection. 
Nevertheless  the  positions  of  the  projections  e*  and  ev  are 
independent  of  the  position  of  the  ground  line. 


110 


DESCRIPTIVE   GEOMETRY         [XIII,  §  124 


124.  The  Intersection  of  Two  Limited  Plane  Surfaces.  Let  it 
be  required  to  find  the  intersection  of  the  triangle  abc,  Fig. 
186,  with  the  plane,  indefinite  in  length,  but  limited  in  width 
by  the  parallel  lines  J  and  K.  The  intersection  can  be  found, 
without  rinding  the  traces  of  either  plane,  by  applying  the 
preceding  method,  as  follows.  Using  the  auxiliary  plane  X 
which  contains  J  and  is  perpendicular  to  H,  we  find  that  the 
line  J  intersects  the  plane  of  the  lines  ab  and  be  in  the  point 
d.  Using  the  auxiliary  plane  Y,  containing  K,  this  line  inter- 
sects the  plane  of  the  lines  ac  and  be  in  the  point  e.  There- 
fore the  plane  JK  intersects  the  plane  abc  in  tlie  line  de. 

125.  Visibility.  If  both  the  plane  surfaces  of  Fig.  186  are 
considered  to  be  opaque,  each  of  them  must  hide  a  portion  of  the 


Fig.  186. 


Fig.  187. 


other.  The  visibility  of  either  projection  must  be  determined 
by  means  of  information  obtained  from  the  other  projection. 
Thus,  to  determine  the  visibility  of  the  Zf-projection,  take  any 
point  in  which  intersect  the  projections  of  two  lines  not  in  the 
same  plane.  For  example,  consider  the  point  where  Kh  inter- 
sects bhch.     This  is  actually  the  projection  of  two  points,  n  in 


XIII,  §  126]         INTERSECTION   OF   PLANES  111 

the  line  be,  and  o  in  K.  Project  to  the  ^projection,  obtaining 
nv  and  o".  From  these  projections  we  see  that  the  point  o  is 
higher  than  the  point  n ;  that  is,  iT  passes  above  be.  Therefore, 
ha  the  iZ-projection,  Kh,  which  contains  oh,  is  the  visible  line 
at  the  point  under  consideration. 

The  other  points  in  the  ^-projection  where  the  projections  of 
lines  not  in  the  same  plane  cross  each  other  can  be  tested  in 
the  same  way.  This  process  finally  results  in  the  visibility 
shown  in  the  figure. 

We  may  reason  also  as  follows  :  the  line  K  has  been  found 
to  be  above  the  triangle  at  the  point  o.  Now  K  intersects  the 
triangle  at  the  point  e.  Therefore  beyond  e  the  line  K  passes 
out  of  sight  under  the  triangle,  so  that  at  the  point  where  Kh 
intersects  ahch,  the  latter  must  be  the  visible  line  ;  and  so  on 
around. 

The  visibility  of  the  P"-projection  is  similarly  determined. 
Begin  at  any  point  where  the  two  projections  of  lines  not  in 
the  same  plane  cross  each  other,  as  for  instance  where  Jv  inter- 
sects avcv.  Project  to  the  JET-view  to  find  which  line  is  in  front 
of  the  other.  In  this  case,  point  r  in  ac  is  in  front  of  point  q 
in  J,  therefore  avcv  is  visible,  and  Jv  is  hidden  by  the  triangle. 
And  so  on  until  the  complete  visibility  is  found. 

126.  The  Intersection  of  Two  Planes  Limited  in  One  Direction. 
Figure  187  represents  the  intersection  of  two  planes,  each 
limited  in  width  but  indefinite  in  length.  This  case  presents 
one  new  feature  over  Fig.  18(>.  Using  auxiliary  planes  per- 
pendicular to  H,  the  line  J  intersects  the  plane  AB  in  the  point 
d.  But  the  line  K  does  not  intersect  the  plane  AB  within  the 
limits  of  its  extent. 

If,  however,  the  plane  AB  extended  indefinitely  beyond  B, 
K  would  intersect  this  plane  in  the  point  e,  outside  of  B,  and 
cle  (d"ev,  dheh)  would  be  the  line  of  intersection  of  the  plane 
.IK  with  such  a  plane.  Therefore,  the  portion  df  of  the  line 
de,  which  lies  within  the  limits  of  the  plane  AB,  must  be  the 
intersection  required. 

The  visibility  of  the  two  projections  is  obtained  as  explained 
above. 


112 


DESCRIPTIVE   GEOMETRY        [XIII,  §  127 


127.  The  Intersection  of  a  Plane  and  a  Pyramid.  Before  be- 
ginning to  find  the  intersection,  visualize  the  solid,  to  deter- 
mine its  visible  faces.  Then  when  the  first  line  of  intersection 
is  found,  if  it  be  on  a  visible  face,  make  it  a  full  line.  If  it  is 
on  an  invisible  face,  make  it  a  dotted  line ;  and  so  on  for  the 
complete  intersection.  The  evident  advantage  of  this  is  that 
the  student  will  be  able  to  visualize  his  work  more  readily  and 
more  completely  as  he  goes  along.     The  construction  of  the 


Fig.  188. 


intersection  requires  merely  the  application  and  extension  of 
the  preceding  method. 

In  Fig.  188  let  it  be  required  to  find  the  intersection  of  the 
limited  plane  JK  with   the  faces  of   the  triangular  pyramid 


XIII,  §  127]        INTERSECTION   OF  SOLIDS  113 

oabc.  In  the  plan,  all  the  faces  are  visible  except  the  base 
abc ;  in  the  elevation  oab  and  obc  are  visible,  while  oac  is  not 
visible.  In  beginning  to  find  the  intersection,  take  the  lines 
J  and  K  separately.  Select  any  face  of  the  pyramid,  as  oab. 
Find  the  point  4  where  J  intersects  the  face  oab,  by  using  the 
plane  X  perpendicular  to  V.  Similarly  find  9,  the  point  where 
K  intersects  the  same  face.  The  line  4-9  is  then  the  inter- 
section of  plane  JK  with  face  oab,  and  is  made  a  full  line  in 
both  views,  as  oab  is  a  visible  face  in  both  plan  and  elevation. 
Taking  another  face  of  the  pyramid,  as  oac,  and  proceeding  as 
before,  K  is  found  to  intersect  at  point  10,  while  J  intersects 
at  the  point  11,  in  the  plane  of  oac,  but  outside  the  face  itself. 
Here  it  is  well  to  note  that  while  the  actual  face  of  the  pyramid 
is  of  limited  extent,  the  plane  of  the  face  is  unlimited.  Joining 
10  with  11,  the  part  from  10  to  12  is  the  actual  intersection  of 
JK  with  the  face  oac,  is  a  full  line  in  plan,  and  dotted  in  ele- 
vation. The  line  J  has  been  found  to  enter  the  pyramid  at 
point  4,  but  the  point  where  it  comes  out  has  not  yet  been 
found.  By  inspection  of  the  plan,  J"  will  necessarily  come  out 
on  face  obc.  The  construction  gives  5  as  the  point  where  J 
pierces  this  face.  The  intersection  of  KJ  with  the  face  obc 
will  be  the  line  5-12,  for  as  12  is  on  the  edge  oc  of  the 
pyramid,  it  must  be  a  point  common  to  the  intersections  on 
the  adjoining  faces  oca  and  ocb.  The  line  5-12  is  visible  in 
both  plan  and  elevation. 

It  should  be  observed  that  since  point  12  is  the  intersection 
of  plane  JK  with  the  edge  of  the  pyramid,  it  might  have  been 
obtained  directly  by  finding  where  oc  intersected  the  plane  JK. 
This  would  have  been  a  convenient  construction  in  case  the 
point  11  luid  fallen  outside  the  limits  of  the  drawing. 

Completive  the  Visibility.  With  the  intersection  itself  prop- 
erly lined  in  with  lull  and  dotted  line,  the  visible  portion  of 
each  edge  of  the  pyramid  or  plane  can  usually  be  determined 
l>v  inspection.  In  case  of  doubt,  however,  the  relative  position 
of  two  edges  which  apparently  intersect  in  one  view  may  be 
determined  l>\  projecting  the  point  of  apparent  intersection  to 
the  other  view  (§  125). 


114  DESCRIPTIVE   GEOMETRY         [XIII,  §  128 

128.  The  Intersection  of  Two  Solids  Bounded  by  Plane  Faces. 
By  further  extending  the  preceding  methods,  the  intersections 
with  each  other  of  the  surfaces  of  any  two  solids  bounded  by 
plane  faces  —  a  broken  line  or  lines  usually  known  as  intersec- 
tion of  the  solids  —  can  be  found.  The  detail  of  the  construc- 
tion varies  in  each  particular  case,  but  involves  nothing  which 
has  not  been  already  explained.  In  order  not  to  get  lost  in  the 
construction,  however,  it  is  necessary  to  proceed  in  a  system- 
atic and  orderly  manner ;  therefore,  for  the  benefit  of  stu- 
dents who  may  wish  to  follow  through  such  a  construction 
before  attempting  one  for  themselves,  we  shall  give  the  entire 
detail  of  one  typical  construction. 

The  two  solids  chosen,  Fig.  189,  are  a  triangular  right  prism 
whose  long  edges  are  the  lines  J,  K,  and  L,  and  a  wedge,  with 
a  rectangular  base  adbe,  and  an  edge  cf  parallel  to  the  plane  of 
the  base.  Before  starting  to  draw,  we  should  notice  from  the 
//-projection  that  the  ends  of  the  prism  JKL  are  not  con- 
cerned in  the  intersection  at  all.  Therefore  it  will  be  well  to 
begin  by  finding  where  the  lines  J,  K,  and  L,  considered  sepa- 
rate^, intersect  the  other  solid.  We  may  notice  also,  if  we 
wish,  that  the  face  def  of  the  wedge  will  not  be  intersected  by 
the  prism.  Again,  it  is  evident  that  the  line  ad  will  not  inter- 
sect the  prism.  This  information,  however,  is  not  of  particular 
importance  at  the  start. 

Before  beginning  to  draw,  it  is  well  also  to  determine  the 
visible  faces  of  each  solid  considered  alone.  Then,  when  any  line 
of  the  intersection  is  found,  it  can  be  visible  only  when  it  is  the 
intersection  of  two  visible  faces,  one  of  each  solid.  The  visi- 
bility of  the  intersection  being  thus  determined,  and  the  visible 
faces  of  each  solid  taken  alone  being  known,  the  visible  lines 
in  the  combined  figure  usually  can  be  determined  by  inspection. 
In  the  case  in  hand,  the  F-projection  shows  that  ad  is  the 
highest  edge  of  the  wedge,  hence  ad  is  visible  in  plan.  There- 
fore tf/and  de  are  also  visible.  It  follows  that  the  faces  adfc, 
dihh.  and  dfe  are  visible  in  plan.  The  plan  shows  cf  as  the 
extreme  back  edge,  hence  in  elevation  this  edge  is  invisible. 
This  means  that  adeb  is  the  only  visible  face  in  this  view.     For 


XIII,  §  128]        INTERSECTION   OF   SOLIDS 


115 


Fia.  189.  —  The  Intersection  of  Two  Solids  Bounded  by  Plane 

Faces. 


116  DESCRIPTIVE   GEOMETRY        [XIII,  §  128 

the  other  solid,  «/,  the  highest  edge,  is  visible  in  plan,  so  the 
faces  JL  and  JK  are  seen  in  plan.  In  elevation,  «/A"and  KL 
are  the  visible  faces. 

To  find  the  actual  intersection.  Choose  any  face  of  the  wedge, 
as  adeb,  and  find  where  all  three  edges  K,  J,  and  L  in  turn 
intersect  this  face.  The  plane  X,  perpendicular  to  H,  passed 
through  K,  intersects  adeb  in  the  line  2-3,  which  crosses  K 
(seen  in  elevation)  in  point  6 ;  K  intersects  adeb  then  in  point  6. 
In  the  same  way  J  is  found  to  intersect  adeb  in  point  12  ;  join- 
ing 6  and  12  we  have  the  intersection  of  face  KJ  with  adeb. 
Since  both  surfaces  are  visible  in  both  plan  and  elevation,  the 
line  6-12  is  a  full  line  in  each  view.  The  line  L  is  found  to 
intersect  adeb  produced  in  point  18.  Join  12-18  and  6-18, 
retaining  only  the  portions  12-19  and  6-23,  and  the  intersec- 
tion of  the  prism  KJL  with  face  adeb  is  complete.  The  line 
12-19  will  be  invisible  in  elevation  and  visible  in  plan,  while 
6-23  will  be  visible  in  elevation  and  invisible  in  plan.  The 
intersection  of  the  two  solids  must  evidently  continue  from  the 
points  23  and  19  on  to  the  adjoining  face  bcfe,  hence  choose 
next  this  face  and  find  where  J,  K,  and  L  intersect.  Plane  X 
through  K  intersects  this  face  in  the  line  4-3,  almost  if  not 
quite  parallel  to  K.  This  shows  that  K,  and  therefore  also  J 
and  L  are  nearly  parallel  to  the  face  bcfe,  so  their  points  of 
intersection  are  inaccessible.  Therefore  choose  next  another 
face,  as  abc.  Line  K  is  found  to  intersect  at  point  5,  and  J  at 
point  16  in  the  face  produced.  Joining  these  points,  the  part 
5-17  is  the  actual  intersection  of  JK  with  the  face,  and  is 
invisible  in  both  views.  Line  L  does  not  actually  intersect  the 
face,  but  will  intersect  the  face  produced.  The  plane  Z  through 
L  intersects  the  edges  ac  and  be  extended  in  the  points  28  and 
27,  and  this  line  when  produced  will  cut  L  in  29.  Manifestly, 
however,  it  would  be  a  very  inaccurate  construction  to  use 
points  so  close  together  as  27  and  28  when  the  line  must  be 
produced  much  beyond  either  point,  and  such  a  construction 
should  not  be  used  without  some  check  on  its  accuracy.  In 
this  case,  such  a  check  is  readily  available.  The  planes  X,  Y, 
and  Z  are  parallel,  and  their  intersections  with  the  plane  abc 


XIII,  §  128]        INTERSECTION   OF   SOLIDS 


117 


Fig.  189.  —  The  Intersection  of  Two  Solids  Bounded  by  Plane 
Faces. 


118  DESCRIPTIVE   GEOMETRY        [XIII,  §  128 

must  be  three  parallel  lines.  Hence,  in  extending  27-28  to  L 
at  29,  care  should  be  taken  to  make  the  line  parallel  to  the 
intersections  1—4  and  7-10.  Lines  K,  J,  and  L  intersect  the 
face  abc  or  the  face  produced  in  points  5,  16,  and  29  respec- 
tively. Joining  these  points,  the  parts  5-17  (already  noted), 
21-22,  and  5-26  form  the  complete  intersection  of  KJL  with 
this  face  of  the  wedge.  From  the  points  17  and  21  the  inter- 
section must  continue  on  to  the  adjoining  face  acfd.  A  little 
reflection  will  show  that  J  must  actually  intersect  this  fare. 
The  construction  determines  the  point  as  11.  As  point  17  is 
in  the  plane  JK  and  also  acfd,  11-17  joined  will  be  the  inter- 
section of  JK  with  face  acfd,  and  likewise  11-21  is  the  intersec- 
tion of  JL  with  the  same  face  of  the  wedge.  There  remains 
to  be  found  the  intersection  on  the  face  bcfe.  No  further  points 
need  to  be  found,  however,  for  as  26  and  23  are  both  in  this 
face  and  also  in  the  plane  KL  the  line  connecting  them  will  be 
the  intersection  of  KL  with  the  face.  Similarly  22-19  is  the 
intersection  of  JL  with  the  face,  which  completes  the  intersec- 
tion of  the  two  solids. 

Having  shown  the  visibility  of  the  intersection  in  both  views 
as  soon  as  found,  notice  that  the  portions  of  the  edges  of  the 
solids  which  are  finally  visible  may  be  readily  determined  by 
inspection.  (The  student  should  take  pains  to  satisfy  himself 
as  to  the  correctness  of  the  visibility  of  each  edge  as  shown.) 

As  a  check,  start  with  any  point  of  the  intersection,  as  5 ; 
by  tracing  5-17-11-21-22-19-12-6-23-26-5,  we  return  to  the 
starting  point.  In  this  case  there  is  but  one  continuous  inter- 
section. With  different  positions  of  the  solids,  however,  there 
may  be  two  intersections.  This  happens,  for  example,  when 
one  solid  completely  penetrates  the  other. 

129.  The  Intersection  of  a  Sphere  with  Another  Solid.  Sup- 
pose that  we  have  two  intersecting  solids,  the  first  a  sphere, 
the  second  any  solid  bounded  by  plane  faces.  Pass  an  auxil- 
iary plane  parallel  to  H,  so  as  to  cut  both  solids.  This  plane 
will  cut  from  the  sphere  a  circle,  and  from  the  other  solid  a 
polygon  of  three  or  more  sides.  Both  the  circle  and  polygon 
will  project  in  true  shape  and  size  in  the  J7-view,  where  any 


XIII,  §  128]        INTERSECTION   OF   SOLIDS 


119 


Fig.  189.  —  The  Intersection  of  Two  Solids  Bounded  by  Plane 
Faces. 


120  DESCRIPTIVE   GEOMETRY        [XIII,  §  129 

points  of  intersection  may  be  noted.  These  points  will  evi- 
dently lie  on  the  line  or  lines  of  intersection  of  the  two  solids. 

An  auxiliary  plane  parallel  to  V  may  be  similarly  used,  the 
sections  cut  from  the  two  solids  appearing  in  true  shape  and 
size  in  the  F-projection.  In  any  particular  problem,  therefore, 
the  solution  would  be  effected  by  planes  parallel  to  H,  or  to  V, 
according  to  convenience.  Indeed,  it  is  often  desirable  to  use 
auxiliary  planes  in  each  of  these  directions. 

The  above  method  of  solution  can  be  applied  to  special  cases 
of  the  intersection  of  a  sphere  with  some  of  the  curved  solids. 
For  example,  let  the  curved  solid  be  a  circular  right  cylinder 
with  its  axis  perpendicular  to  H.  Planes  parallel  to  H  will 
cut  circles,  while  planes  parallel  to  V  will  cut  rectangles  from 
this  cylinder.  Hence  the  analysis  previously  given  can  be 
applied  to  this  case. 

In  general,  if  two  intersecting  solids  are  such,  and  so  placed 
that  planes  parallel  to  H  or  to  V  will  intersect  both  solids  in 
simple,  readily  determined  figures,  such  as  straight  lines  or 
circles,  the  intersection  of  the  solids  may  be  found  by  the  use 
of  such  planes  as  auxiliaries. 

A  few  simple  examples  are  given  in  the  following  Articles. 

130.  The  Intersection  of  a  Sphere  with  a  Prism  (Fig.  190). 
Given  the  sphere  and  the  triangular  right  prism  abc-def,  both 
placed  in  the  first  quadrant.  It  is  evident  that  the  intersection 
is  formed  by  the  three  vertical  sides  of  the  prism,  therefore 
the  plan  of  the  intersection  coincides  with  the  plan  of  the 
prism.  The  intersection  of  each  vertical  side  of  the  prism  is 
here  found  as  explained  in  §  84.     (See  Figs.  108  and  109.) 

Let  us  consider,  for  example,  the  face  cbfd.  The  points  1 
and  5  are  located  at  once  in  the  elevation  on  jB",  the  other  pro- 
jection of  circle  B.  The  plane  R  cuts  from  the  sphere  the 
circle  R.  How  is  its  diameter  found?  The  vertical  side  of 
the.  prism  intersects  this  circle  in  the  points  2  and  8. 

The  other  points,  3,  4,  6,  7,  are  found  in  a  similar  manner, 
and  the  ellipse  drawn.  It  may  be  noted  that  the  ellipse  in 
elevation  has  its  major  axis  vertical  and  equal  in  length  to 
l*-5*,  while  its  minor  axis  is  equal  to  1V-5V. 


XIII,  §  130]         INTERSECTION   OF   SOLIDS 


121 


The  intersections  of  the  other  two  sides  of  the  prism  are 
found  in  a  similar  manner,  as  in  Fig.  190.     The  F"-projection  of 
the  first  intersection  is  a  com- 
plete ellipse,  but  the  other  two 
are    partial    ones,    with    the 
points  10  and  11  in  common. 

Visibility.  In  any  case  of 
the  intersection  of  two  sur- 
faces, not  only  the  visible  part 
of  the  intersection  should  be 
shown,  but  also  the  visible 
edges  and  the  outlines  of  the 
surfaces,  as  in  §  129. 

Visible  Inteksections. 
The  face  first  considered,  cbdf, 
is  the  back  of  the  prism,  hence 
its  intersection  with  the  sphere 
is  wholly  invisible  in  eleva- 
tion. The  other  two  faces  are 
each  visible  in  elevation.  In 
each  case  as  much  of  the  inter- 
section as  lies  on  the  front  or 
visible  half  of  the  sphere  will 
be  visible  in  elevation,  that  is, 

10  to  13  and  11  to  14  on  the 
right-hand  curve,  10  to  16  and 

11  to  17  on  the  left-hand  curve. 
Visible    Edges    axd    Out- 
lines.    The   edge   ae   of   the 

prism  is  the  front  edge,  and  intersects  the  sphere  at  the  points 
10  and  11  on  the  front  half.  Hence  ae  is  visible  from  av  to  10", 
and  from  11"  to  e".  Since  the  edge  cd  is  tangent  to  the  sphere 
at  point  5  on  the  back  half,  cd  disappears  on  V  where  it  crosses 
the  outline  of  the  sphere.  Hence  the  outline  of  the  sphere 
is  visible  from  the  left  toward  the  right  as  far  as  points  16" 
and  17".  Likewise,  the  edge  bf  passes  behind  the  sphere,  and 
the  outline  of  the  sphere  is  visible  from  13*  around  to  14". 


d  c 


Fig.  190. 


122 


DESCRIPTIVE  GEOMETRY        [XIII,  §  131 


131.    The  Intersection  of  a  Sphere  with  a  Right  Cylinder. 

Example  1  (Fig.  191).  Given  a  sphere  and  a  circular  cylin- 
der. The  plan  of  the  intersection  is  evidently  the  arc  of  the 
circle  1*5*7*,  and  a  study  of  this  plan  view  will  show  that  the 
intersection  consists  of  one  continuous  curve.  The  points  1 
and  7,  shown  in  plan  on  the  outline  of  the  sphere,  are  projected 
directly  to  1"  and  7*.     The  points  4,  10,  6,  8,  must  lie  on  the 


Fig.  191. 


Fig.  192. 


great  circle  of  the  sphere  in  elevation,  and  are  projected  at  once 
to  4",  10",  6"  and  8".  Other  points  are  found  by  means  of  aux- 
iliary planes  parallel  to  V,  as  in  the  preceding  figure.  The 
plane  T,  passed  through  ch,  gives  the  points  3"  and  11",  in  which 


XIII,  §  131] 


INTERSECTION    OF   SOLIDS 


123 


the  intersection  is  tangent  to  the  outline  of  the  cylinder.  The 
visibility  of  the  curve  and  of  the  outlines  of  the  surfaces  is 
determined  as  previously  explained. 

Example  2  (Fig.  192).  Given  a  sphere  and  an  elliptical  cylin- 
der. The  intersection  and  the  visibility  are  found  as  in  Ex- 
ample 1.     Notice  that  here  two  parts  of  the  curve  are  visible. 

Example  3  (Fig.  193).  Given  a  sphere  and  a  circular  cylin- 
der. In  this  case  an  inspection  of  the  plan  serves  to  show  that 
the  intersection  will  consist  of  two  curves,  each  entirely  distinct 
from  the  other.  Each  curve  in  elevation  will  be  symmetrical 
with  respect  to  the  horizontal  center  line  of  the  sphere  (why'.'). 
The  solution  is  left  to  the  student. 


Fig.  193. 


Vir,.  mi. 


Example  4  (Fig.  194).  Given  a  sphere  and  a  circular  cylin- 
der. In  this  example  it  may  not  be  so  easy  to  vizualize  the 
intersection  by  considering  the  plan  view.  Nevertheless  a 
little  Btudy  will  show  thai  there  must  be  two  closed  figures 
having  a  common  point  a.  Asa  matter  of  fad  flic  int.!  lection 
when  found  in  elevation  will  he  shaped  like  an  irregular  figure 


124 


DESCRIPTIVE  GEOMETRY        [XIII,  §  131 


8,  the  curve  crossing  itself  (not  tangent),  at  a.     It  is  called 
a  one-curve  intersection.     The  solution  is  left  to  the  student. 

132.  The  Intersection  of  a  Cylinder  and  a  Torus.  Another 
example  is  shown  in  Fig.  195.  An  auxiliary  plane,  as  Q,  par- 
allel to  H,  cuts  the  cylinder  in  the  elements  A  and  B,  and  the 
torus  in  the  circles  C  and  D,  giving  six  points,  1,  2,  ...  6,  of 
the  intersection.    The  chosen  secant  planes,  parallel  to  H,  should 


Fig.  195. 


include  one  through  the  contour  element  E  of  the  cylinder ; 
one  through  the  highest  element,  G ;  one  through  the  center  of 
the  torus ;  and  one  tangent  at  the  bottom  of  the  torus.  The 
points  7  and  8  of  the  intersection  evidently  lie  in  the  plane  X, 
which  passes  through  the  center  of  the  torus  parallel  to  V. 


CHAPTER   XIV 

PROBLEMS    INVOLVING    THE    REVOLUTION   OF    PLANES 

133.  A  Line  Lying  in  a  Given  Plane.  One  method  of  locating 
a  line  so  that  it  will  lie  in  a  given  plane  is  to  assume  the  traces 
of  the  line  in  the  corresponding  traces  of  the  plane.  This  has 
already  been  explained  in  §  97.     (See  Fig.  123.) 


Fig.  123  (repeated). 


Another  method  is  to  assume  one  of  the  projections  of  the 
line.  As  soon  as  either  the  H-  or  the  F-projection  of  a  line  is 
assumed,  the  line  becomes  definite  from  the  condition  that  it 
shall  lie  in  the  given  plane. 

Problem  16.  Given  one  projection  of  a  line  lying  in  a  plane,  to 
find  the  other  projection. 

Analysis.  The  other  projection  of  the  line  will  be  deter- 
mined if  two  points  on  it  are  found.  Whenever  it  is  practi- 
cable, the  simplest  points  to  consider  are  the  two  traces,  s  and  t, 
which  lie,  respectively,  in  the  horizontal  and  vertical  traces  of 
the  given  planes  (§  90).  One  trace  of  the  line  will  be  at  the 
intersection  of  the  given  projection  with  the  corresponding 
trace  of  the  plane,  the  other  projection  of  this  point  being  in 
GL.  The  other  trace  of  the  line  will  be  in  the  other  trace  of 
the  plane,  and  is  determined  by  a  projector  drawn  from  the 
intersection  of  the  given  projection  with  GL. 

125 


126 


DESCRIPTIVE  GEOMETRY 


[XIV,  §  133 


Construction  (Fig.  196).  Let  the  plane  Q  and  the  projection 
Ah  be  given.  The  intersection  of  Ah  and  HQ  is  the  if-trace  of 
the  line  A  (§  96).  The  C-projection,  sv,  of  this  point  lies  in 
GL.     The  intersection  of  Ah  and  GL  is  the  .ff-projection  of  the 


Fig.  196. 


F-trace  of  the  line  (Prob.  1,  §  37).  The  actual  trace,  f,  lies  in 
VQ.  The  projection  A"  is  now  determined,  since  two  points, 
sv  and  f,  are  known. 

Similarly,  the  projection  Ah  may  be  found  if  Av  is  given. 


Fig.  19 


A  second  example  is  given  in  Fig.  197.  The  lettering  and 
explanation  are  the  same  as  for  Fig.  196. 

Special  Case  I.  Suppose  that  the  given  projection  is  par- 
allel (a)  to  the  corresponding  trace  of  the  plane ;  or  (b)  to  the 
ground  line.  In  either  event  the  line  should  be  recognized  as 
one  of  the  principal  lines  of  the  plane  (§  99),  parallel  to  H  or 
V  as  the  case  may  be.  Thus,  in  Fig.  198,  if  Ah  is  given  parallel 
to  HQ,  Av  is  parallel  to  GL ;  or  if  Av  is  given  parallel  to  GL, 
then  Ah  is  parallel  to  HQ.    The  line  A  in  this  case  is  a  horizontal 


XIV,  §  133] 


A  LINE  IX  A  PLANE 


127 


principal  line  of  Q,  having  its  F-trace  at  the  point  t  (th,  t"). 
Similarly,  if  Av  is  given  parallel  to  VQ,  Ah  is  parallel  to  GL, 
and  vice  versa,  the  line  being  a  vertical  principal  line  of  Q. 


Fig.  198. 

Special  Case  II.  Suppose  that  both  the  plane  and  the 
given  projection  of  the  line  are  parallel  to  the  ground  line. 
Whichever  projection  of  the  line  is  given,  the  other  projection 
will  also  be  parallel  to  the  ground  line.  Let  Q,  Fig.  199,  be  the 
given  plane,  and  suppose  Ah  to  be  given.     One  point  will  be 


0. 

Ah 

h              \ 
u 

\               i 

Av 

VQ. 

/"' 

a. 
> 

F 

G.  1!)9. 

sufficient  to  locate  A\  This  point  may  suitably  be  the  profile 
trace  (§  59)  of  the  line,  on  any  assumed  profile  plane,  as  P. 
Find  first  the  profile  trace,  PQ,  of  the  given  plane  (§  60).  Then 
from  w*  we  can  project  to  u"  on  PQ,  thence  to  av  on  VP.     The 


projection  A"  passes  through  //". 
tion  will  locate  Ah  if  A*  is  given. 


The  reverse  (if  this  construe- 


128 


DESCRIPTIVE   GEOMETRY         [XIV,  §  133 


Special  Case  III.  Suppose  that  the  given  plane  is  perpen- 
dicular to  H  or  V.  Let  the  given  plane  Q,  Fig.  200,  be  perpen- 
dicular to  H.  Then  if  Av  is  given,  in  any  position  except 
perpendicular  to  GL,  Ah  coincides  with  HQ,  since  this  trace  is 
an  edge  view  of  the  plane  Q  (§§  51  and  98).  In  the  exceptional 
position,  shown  by  B°  perpendicular  to  GL,  the  .//-projection, 
Bh,  reduces  to  a  point  on  HQ. 


Fig.  200. 


Fig.  201. 


Conversely,  if  Ah  is  given  coincident  with  HQ,  Av  may  be 
taken  at  random  in  any  position  except  perpendicular  to  GL, 
and  the  case  is  indeterminate. 

But  if  the  given  plane  Q,  Fig.  201,  be  perpendicular  to  H,  and 
if  Ah  be  assumed  not  coincident  with  HQ,  the  solution  becomes 
impossible,  for  there  are  no  points  in  the  plane  Q  to  correspond 
with  the  assumed  projections  ch,  dh,  or  in  fact  with  any  point 
of  Ah  except  the  intersection,  eh,  of  Ah  and  HQ.  The  results 
are  similar  if  the  given  plane  is  perpendicular  to  V. 

Corollary  I.  Given  one  projection  of  a  point  lying  in  a 
plane,  to  find  the  other  projection. 

Analysis.  Through  the  given  point  pass  any  line  lying  in 
the  plane.  Find  the  other  projection  of  the  line.  This  pro- 
jection must  contain  the  required  projection  of  the  given  point. 

Construction  (Fig.  202).  Let  Q  be  the  given  plane,  and  asr 
sume  ah  to  be  given ;  it  is  required  to  find  a".  Through  a* 
draw  the  //-projection,  Mh,  of  some  line  in  the  plane  Q.  In 
general,  the  simplest  construction  results  when  the  assumed 
line  is  a  principal  line  of  the  plane  ;  hence  Mh  is  here  drawn 
parallel  to  HQ.     Find  M v  (Special  Case  I).     Then  u"  lies  on 


XIV,  §  133] 


A  LINE  IN  A  PLANE 


129 


Mv  and  is  obtained  by  projecting  from  ah.    The  reverse  of  this 
construction  will  locate  a*  when  a"  is  given. 


Fig.  202. 


Fig.  203. 


Corollary  II.  To  find  the  second  projection  of  a  line  lying 
in  a  plane  when  the  general  solution  fails,  partially  or  wholly. 

Let  Q,  Fig.  203,  be  the  given  plane,  and  let  ah  bh  be  given  so 
that  it  cannot  be  produced  to  intersect  either  HQ  or  GL 
within  the  limits  of  the  figure.  Then  a"  and  bv,  or  in  general 
any  two  points  on  the  line,  may  be  located  by  Corollary  I. 

Corollary  III.  To  find  a  line  of  maximum  inclination  of 
a  plane. 

Let  Q,  Fig.  204,  be  the  given  plane.  A  line  of  maximum 
inclination  to  77  is  perpendicular  to  HQ  (§  111).  Hence  assume 
Ah  perpendicular  to  HQ,  and  hnd  Av  by  the  general  method. 


Fig.  201. 


Fig.  205. 


In  Fig.  205,  the  line  7>  is  in  the  plane  R,  and  is  a  line  of 
maximum  inclination  to  V.  The  projection  11"  is  taken  per- 
pendicular to  VR,  and  then  Bh  is  found. 

K 


130 


DESCRIPTIVE  GEOMETRY 


[XIV,  §  134 


134.    Revolution  of  a  Plane  About  an  Axis  Perpendicular  to  H 

or  V.  As  an  illustration  of  the  general  case  of  this  problem, 
let  it  be  required  to  revolve  the  plane  Q,  Fig.  206,  through  an 
angle,  «,  about  the  line  A,  perpendicular  to  H,  as  an  axis. 
The  axis  pierces  Q  at  the  point  c  (cA,  c").  This  point  is  found 
most  readily  by  observing  that  since  Ah  is  a  point,  ch  coincides 
with  Ah.  Then  ch  is  one  projection  of  a  point  lying  in  plane 
Q,  hence  c°  is  the  corresponding  projection  (Prob.  16,  Cor.  1, 
§  133).  In  any  revolution  of  Q  about  A,  the  point  c  will  re- 
main fixed.  Now  a  plane  is  determined  when  a  point  and  a 
line  are  known ;  hence,  having  the  fixed  point  c,  it  will  be 
necessary  to  revolve  but  one  line  of  Q.  This  line  may  con- 
veniently be  the  //-trace,  IIQ,  of  Q,  which  is  now  revolved 
about  ch  as  a  center  to  the  required  position  HR.  The  F-trace, 
VR,  is  then  determined  by  the  fact  that  the  plane  R  must  con- 
tain the  point  c.  Consequently  R  contains  the  line  N,  drawn 
through  c  and  parallel  to  HR.     (See  Example  3,  §  108.) 

Let  now  the  plane  Q,  Fig.  206,  be  further  revolved,  until  the 
//-trace  takes  position  HS,  perpendicular  to  the  ground  line. 


Fig.  206. 


Fig.  207. 


In  this  new  position,  the  plane  will  be  perpendicular  to  V. 
Hence  the  F-trace,  VS,  will  be  an  edge  view  of  the  plane,  and 
must  therefore  pass  directly  through  c". 

A  plane  may  be  revolved  until  perpendicular  to  H  by  using 


XIV,  §  135]     THE  REVOLUTION  OF  PLANES 


131 


an  axis  perpendicular  to  V.  This  is  done  in  Fig.  207,  where 
the  plane  Q  is  revolved  about  the  line  A,  perpendicular  to  V, 
until  the  P"-trace  of  the  plane,  in  its  new  position  VR,  is 
perpendicular  to  the  ground  line.  Then  HR  is  an  edge 
view  of  the  plane,  and  passes  through  c\  In  this  figure,  the 
axis  A,  besides  being  perpendicular  to  V,  is  taken  lying 
in  H}  which  simplifies  the  hnding  of  the  point  c,  in  which  A 
intersects  Q. 

As  in  the  revolution  of  a  straight  line  (§§  78,  79),  a  plane, 
when  revolved  about  an  axis  perpendicular  to  H,  maintains 
always  its  original  angle  with  H.  If  the  axis  is  perpendicular 
to  V,  the  angle  between  the  revolving  plane  and  V  does  not 
change. 

135.  The  Distance  between  Two  Parallel  Planes.  As  a  simple 
application  of  the  revolution  of  planes  about  axes  perpendicular 
to  H  or  V,  we  will  consider  the  following  Problem. 

Problem  17.  To  find  the  perpendicular  distance  between  two 
parallel  planes. 

Analysis.  Assume  an  axis  perpendicular  to  H  (or  V),  and 
revolve  both  planes  about  this  axis  until  they  are  perpendicular 
to  V(oxH).  The  perpendicu- 
lar distance  between  the  planes 
will  then  appear. 

Construction  (Fig.  208).  Let 
Q  and  R  be  the  given  planes. 
Assume  the  axis  A  perpendicu- 
lar to  H,  and  for  convenience 
lying  in  V.  Revolve  both 
planes  about  the  axis  A  until 
they  are  perpendicular  to  V 
(§  134).  Then  the  new  F-traces, 
VQi  and  VRU  are  the  edge 
views  of  the  planes,  and  the 
perpendicular  distance,  p,   be-  FlG'  208' 

tween  these  traces  is  equal  to  the  distance  between  the 
planes. 


132  DESCRIPTIVE   GEOMETRY         [XIV,  §  135 

Corollary.  To  find  the  perpendicular  distance  from  a  point 
to  a  plane. 

Analysis.  Take  an  axis  perpendicular  to  H  (or  V)  through 
the  given  point.  About  this  axis,  revolve  the  plane  until  it  is 
perpendicular  to  V  (or  H).  Then,  since  the  given  point  on 
the  axis  does  not  move,  or  change  its  relation  to  the  plane, 
the  perpendicular  distance  appears  directly  in  the  V  (or  H) 
projection. 

The  construction  is  left  to  the  student.  The  problem  is  the 
same  as  Problem  14  (§  120),  a  different  method  of  solution  hav- 
ing now  been  found. 

136.  The  Angles  between  an  Oblique  Plane  and  the  Coordinate 
Planes.  The  angle  which  a  given  plane  Q,  oblique  to  both  H 
and  V,  makes  with  either  coordinate  plane,  for  example  H,  may 
be  found  : 

(a)  by  drawing  in  the  plane  a  line  of  maximum  inclination 
to  H  (Prob.  16,  Cor.  3,  §  133)  and  then  finding  the  angle  which 
this  line  makes  with  H  (§  79  or  §  82); 

(6)  by  finding  the  edge  view  of  the  plane  on  a  secondary 
plane  of  projection  taken  perpendicular  to  HQ  (§  70) ; 

(c)  by  revolving  the  plane  about  an  axis  perpendicular  to  H 
until  the  plane  is  perpendicular  to  V  (§  134). 

The  various  constructions  resulting  from  these  different 
methods  resemble  each  other  to  a  considerable  extent,  and  are 
essentially  the  same,  so  that  it  is  needless  to  take  them  all  up 
in  detail.  The  method  chosen  in  this  text  is  thought  to  be  the 
one  which  best  shows  the  connection  between  this  problem  and 
the  two  that  follow. 

Problem  18.  To  find  the  angles  which  an  oblique  plane  makes 
with  H  and  V. 

Analysis.  To  find  the  angle  between  the  given  plane  and  H, 
assume  an  axis  of  revolution  lying  in  P"and  perpendicular  to 
H.  Revolve  the  plane  until  perpendicular  to  V,  when  the 
angle  between  the  new  F-trace  and  the  ground  line  will  be  the 
angle  the  plane  makes  with  H  (§§  134,  136).  Similarly  for 
the  angle  between  the  given  plane  and  V. 


XIV,  §  136]     THE  REVOLUTION  OF  PLANES 


133 


Construction  (Fig.  209).  Let  Q  be  the  given  plane.  Assume 
the  axis  A,  lying  in  V  and  perpendicular  to  H.  Revolve  Q 
about  A  to  the  position  R  (HR,  VR), 
perpendicular  to  V  (§  134).  Then 
the  angle  between  VR  and  GL  is 
the  angle  a,  which  Q  makes  with  H. 
Similarly,  by  revolving  about  the 
axis  B,  lying  in  H  and  perpendicular 
to  V,  we  obtain  the  angle  (3,  which  Q 
makes  with  V. 

It  will  be  seen  in  Fig.  209  that  the 
traces  HR  and  VS,  perpendicular  to 
GL,  are  not  essential  to  the  required 
angles  «  and  /3,  and  that  these  traces 
may  therefore  be  omitted.  This  is 
done  in  Fig.  210,  in  which  are  found 
the  angles  a  and  fi,  which  the  given  plane  Q  makes  with 
H  and  V,  respectively. 


Fig.  209. 


Fig.  210. 


Q. 

I 

Ha 

\ 

\ 
\ 

*     / 

a    / 

Y       va 

Q. 
> 

Fig. 

211. 

Special  Case.  Let  the  given  plane  Q,  Fig.  211,  be  parallel 
to  the  ground  line.  Find  the  profile  trace  of  ^>on  any  assumed 
profile  plane.  This  trace,  PQ,  is  an  edge  view  of  the  plane 
(S  60),  and  gives  the  required  angles,  a  with  //and  (3  with  V, 
directly. 


134 


DESCRIPTIVE   GEOMETRY         [XIV,  §  137 


137.  Planes  making  Given  Angles  with  H  and  V.  The  pre- 
ceding problem  gives  rise  to  two  converse  problems,  in  which 
it  is  required  to  find  planes  which  make  given  angles  with  the 
coordinate  planes. 

Problem  19.  Given  one  trace  of  a  plane,  and  the  angle  which  the 
plane  makes  with  either  H  or  V,  to  find  the  other  trace  of  the  plane. 
There  are  two  cases,  which  are  stated  below. 

Analysis.  The  required  result  may  be  obtained  by  revers- 
ing the  construction  of  Problem  18. 

Case  I.  Given  one  trace  and  the  angle  which  the  plane  makes 
ivith  the  same  coordinate  plane. 

Construction  (Fig.  212).  Let  HQ  and  the  angle  «,  which  Q 
makes  with  H,  be  given.  Assume  the  axis  A,  lying  in  P~and 
perpendicular  to  II.  Revolve  the  given  trace  HQ  about  this 
axis  to  the  position  IIR,  perpendicular  to  GL.  Complete  the 
plane  in  this  position  by  drawing  VR  at  the  given  angle,  a, 


Fig.  212.  Fig.  213. 

with  H.  Then  R  intersects  the  axis  A  at  the  point  c  (C,  c*). 
Let  R  now  be  supposed  to  be  revolved  back  about  the  axis  A 
until  IIR  coincides  with  HQ  ;  the  point  c  remains  fixed.  Since 
c"  is  in  V,  VQ  must  pass  through  c",  and  is  determined  by  this 
point  and  the  point  in  which  HQ  intersects  GL. 

The  given  trace  HQ  may  also  be  revolved  to  the  position 
HRU  giving  a  second  point  <y\  and  a  second  result,  VQV  The 
distances  of  cv  and  Cj"  from  GL  are  obviously  the  same,  hence 
VQ  and  VQt  make  equal  angles  with  GL. 


XIV,  §  137]     THE  REVOLUTION  OF  PLANES  135 

It  is  evident  in  Fig.  212  that  the  construction  is  possible  for 
any  given  angle  a  between  0°  and  90°.  If  a  =  90°,  but  one 
result  is  possible,  and  no  construction  is  necessary.     Why  ? 

Case  II.  Given  one  trace  and  the  angle  which  the  plane 
makes  with  the  other  coordinate  plane. 

Construction  (Fig.  213).  Let  VQ  and  the  angle  «,  which  the 
plane  makes  with  H,  be  given.  Assume  an  axis,  A,  lying  in 
V  and  perpendicular  to  H.  Since  Av  and  VQ  are  both  lines  in 
V,  they  intersect,  locating  the  point  c  (C,  ch),  in  which  A  inter- 
sects the  plane  Q.  Place  the  plane  R,  perpendicular  to  V,  so 
as  to  pass  through  c,  and  making  the  given  angle,  a,  with  H. 
Revolve  R  about  A.  In  this  revolution,  the  angle  a  with  H 
does  not  change  (§  134) ;  hence  HQ  must  be  tangent  to  the 
arc  ef,  drawn  with  ch  as  center  and  radius  che.  This,  together 
with  the  fact  that  HQ  must  pass  through  the  point  n,  in 
which  VQ  intersects  GL,  determines  HQ. 

The  point  e  may  also  be  revolved  in  the  opposite  direction, 
along  the  arc  ef,  thus  giving  a  second  possible  result,  HQX, 
drawn  through  n  and  tangent  to  ef.  It  is  evident  that  HQ 
and  HQl  make  equal  angles  with  the  ground  line. 

This  construction  is  possible  only  when  the  given  acute 
angle  «,  which  Q  makes  with  H,  is  greater  than  the  acute 
angle  which  VQ  makes  with  GL.  If  «  is  given  as  exactly 
equal  to  the  angle  between  VQ  and  GL,  no  construction  is 
necessary,  and  only  one  result  is  possible.     Why  ? 

Special  Case.  The  two  cases  of  this  problem  merge  into 
one  when  the  given  trace  is  parallel  to  the  ground  line,  the 
solution  being  effected  by  means  of  an  assumed  profile  plane. 
(Compare  Fig.  211.)  There  are  always  two  possible  results 
for  any  given  value  of  a  or  ft  greater  than  0°  and  less  than  90°. 

Problem  20.  Given  the  angles  which  a  plane  makes  with  H  and 
V,  to  find  the  traces  of  the  plane. 

As  in  the  preceding  problems,  we  shall  call  the  angle  with 
//.  u,  and  the  angle  with  V,  ft.  If  any  plane  can  be  found 
which  makes  the  angles  a  with  //  and  ft  with  V,  there  is  an 
infinite  number  of  parallel  planes  making  the  same  angles.    The 


136 


DESCRIPTIVE  GEOMETRY         [XIV,  §  137 


location  of  the  plane  becomes  definite  only  when  some  point 
lying  in  the  plane  is  known.  But  a  plane  may  be  readily  trans- 
ferred, parallel  to  itself,  from  one  position  to  another  (see  Prob. 
9,  §  107).  Hence,  the  problem  as  stated  will  be  considered 
solved  when  any  one  of  the  series  of  parallel  planes  is  found. 

Analysis.  It  is  evident  that  the  construction  in  Fig.  209,  in 
which  «  and  yS  are  found  for  a  given  plane  Q,  cannot  be  readily 
reversed.  But  consider  Fig.  214,  in  which  a  and  ft  for  the  given 
plane  Q  are  found  by  taking  the  axes 
of  revolution,  A  and  B,  through  the 
same  point,  o,  in  the  ground  line. 
Since  VR  is  an  edge  view  of  Q,  the 
perpendicular  distance  from  o  to  Q 
is  equal  to  the  perpendicular  ok, 
dropped  from  o  to  VR.  Moreover, 
since  HS  is  an  edge  view  of  Q,  the 
perpendicular  distance  from  o  to  Q 
is  equal  to  the  perpendicular  om, 
dropped  from  o  to  HS.  Hence  ok  = 
om,  whence  it  follows  that  the  two 
edge  views,  VR  and  HS,  are  each 
tangent  to  a  circle  drawn  with  o  as 
center  and  radius  ok  =  om.  The 
plane  Q  may  now  be  obtained  from  the  angles  a  and  /?  by 
reversing  this  construction. 

Construction  (Fig.  215).  Assume  any  point  o  on  the  ground 
line.  Through  o  draw  an  axis  perpendicular  to  the  ground 
line.  AVith  o  as  center,  draw  a  circle  with  any  assumed 
radius,  the  radius  being  the  distance  from  o  to  the  required 
plane  Q. 

Tangent  to  the  circle,  draw  the  edge  view  VR,  which  repre- 
sents the  plane  perpendicular  to  V,  at  the  given  angle  a  with 
H  Xow  VR  intersects  the  axis  at  c,  and  the  ground  line  at  e. 
Hence  c  is  the  fixed  point  on  VQ,  while  HQ  must  be  tangent 
to  the  arc  ef,  drawn  about  o  as  center  with  the  radius  oe. 

Again,  tangent  to  the  circle,  draw  the  edge  view  HS,  which 
represents  the  plane  perpendicular  to  H,  at  the  given  angle  /3 


Fig.  214. 


XIV,  §  137]     THE  REVOLUTION  OF  PLANES 


137 


with  V.  We  find  that  HS  intersects  the  axis  at  d,  and  the 
ground  line  at  g.  Hence,  d  is  the  fixed  point  on  HQ,  while  VQ 
must  be  tangent  to  the  arc  gj,  drawn  with  center  o  and  radius  og. 


Fig.  215. 


We  now  have  one  point  each  on  HQ  and  VQ,  and  an  arc  to 
which  each  is  tangent.     Therefore  HQ  and  VQ  are  determined. 

Check.  Since  HQ  and  VQ  are  independently  determined, 
they  must  intersect  the  ground  line  at  the  same  point,  n. 

If  «  and  /3  are  given  as  acute  angles,  as  is  usual,  a  solution 
is  possible  only  when  the  sum  of  a  and  /3  is  not  less  than  90° 
nor  more  than  180°.  P>ut  when  the  sum  of  these  angles  is 
greater  than  90°,  and  each  angle  is  less  than  90°,  four  possible 
non-parallel  results  may  be  found. 

Thus,  having  found  the  plane  Q,  Fig.  215,  making  a  with  // 
and  (3  with  V,  make  on'  =  on.  Then  it  is  evident  by  symme- 
try that  the  plane  T,  drawn  with  VT  through  c  and  n',  and 
HT  through  (I  and  //',  makes  the  same  angles.  Also,  make 
oc'  =  oc,  and  draw  VX  through  n  and  c'.  Then  VX  makes  the 
same  angle  with  d L  as  does  VQ,  hence  taking  HX  coincident 
with  HQ,  plane  A'  makes  with  //  and  V  the  same  angles  as 
does  Q  (compare  Fig.  212).  A  fourth  result,  plane  Y  (VY, 
HY)  is  obtained  by  symmetry  from  plane  X. 


138 


DESCRIPTIVE   GEOMETRY         [XIV,  §  137 


There  are  other  methods  of  obtaining  the  three  additional 
results  after  the  first  plane  Q  is  obtained.  Thus,  a  distance 
od'  =  ocl  might  be  laid  off  below  the  ground  line,  and  //-traces 
drawn  from  d'  to  n  and  n'.  P>ut  after  four  non-parallel  results 
are  obtained,  all  further  results  will  be  parallel  to  some  plane 
already  found.  This  is  because  there  are  but  four  possible 
slopes  of  an  oblique  plane  :  (1)  downward,  forward,  to  the 
right ;  (2)  downward,  backward,  to  the  right ;  (3)  downward, 
forward,  to  the  left ;  (4)  downward,  backward,  to  the  left. 
(See  §  49.)  The  slope  of  a  plane  is  the  same  as  the  slope  of 
its  lines  of  maximum  inclination  to  H.  These  lines,  if  desired, 
may  be  found  by  Problem  16,  Cor.  3,  §  133. 

Special  Case.  The  sum  of  a  and  ft  is  90°.  The  plane  is 
then  parallel  to  the  ground  line,  and  its  profile  trace  is  its  edge 
view.  Hence  (Eig.  216),  assume  any  profile  plane  of  projec- 
tion. Draw  two  non-par- 
allel profile  traces  making 
the  given  angles  with  H 
and  V,  and  find  the  corre- 
sponding H-  and  F"-traces. 
Only  two  results  are  possi- 
ble, corresponding  to  the 
two  possible  slopes  :  down- 
ward and  forward,  and 
downward  and  backward 
(or  upward  and  backward, 
upward  and  forward). 

Other  special  cases, 
which  may  be  solved  by 
inspection,  result  when 
(1)  either  a  or  ft  is  0°,  in  which  case  the  other  angle  must  be 
90°  (one  result)  ;  (2)  either  «  or  ft  is  90°,  the  other  angle  being 
greater  than  0°  and  less  than  90°  (two  results)  ;  (3)  a  =  ft  =  90° 
(one  result).     These  cases  are  left  to  the  student. 


Fig.  216. 


CHAPTER   XV 

OTHER  PROBLEMS  INVOLVING  THE  REVOLUTION  OF 
PLANES 

138.  The  Revolution  of  a  Plane  about  One  of  Its  Traces.  A 
very  useful  revolution  of  a  plane  is  that  in  which  the  plane  is 
revolved  about  one  of  its  own  traces  as  an  axis.  For  example, 
the  plane  Q  may  be  revolved  about  HQ  as  an  axis  until  Q 
coincides  with  H.  The  figure  is  left  for  the  student  to  draw. 
Any  points,  lines,  or  figures  contained  in  Q  Avill  be  carried  by 
this  revolution  into  the  coordinate  plane.,  where  they  will 
appear  in  their  true  sizes  and  relations. 

Problem  21.  To  find  the  position  of  a  point  lying  in  a  plane, 
when  the  plane  is  revolved  into  H  or  V  about  the  corresponding  trace. 

Note.  Since  only  one  projection  of  a  point  lying  in  a  plane  can,  in 
general,  be  assumed,  it  may  be  necessary  to  find  the  second  projection  by 
tbe  use  of  Problem  10  (§133),  before  attempting  to  solve  the  present 
X>roblem. 

Analysis.  Suppose  point  a  lying  in  plane  Q,  to  be  revolved 
about  IIQ  until  it  lies  in  //.  The  figure  is  left  to  the  student. 
The  process  is  the  same  as  described  in  §  70,  and  shown  pic- 
torially  in  Fig.  08.  The  statement  of  §  76  may  be  modified  to 
meet  the  present  situation,  as  follows.  If  a  point  a  Lying  in 
plane  Q,  be  revolved  about  IIQ  as  an  axis,  the  circular  path 
of  the  revolving  point  will  project  on  //as  a  straight  line  per- 
pendicular to  IIQ.  If  the  point  a  be  revolved  into  //,  its 
revolved  position  will  be  at  a  distance  from  H(J  equal  to  the 
true  distance  of  the  point  from  HQ. 

An  analogous  statement  applies  to  the  revolution  of  a  point 
about  VQ  into  V. 

139 


140 


DESCRIPTIVE  GEOMETRY 


[XV,  §  138 


Construction  (Fig.  217).  Let  a,  lying  in  the  plane  Q,  be  the 
given  point,  and  let  it  be  required  to  revolve  a  about  HQ  into 
H  Erom  ah  draw  the  line  aheh  perpendicular  to  HQ,  and  pro- 
duce the  line  indefinitely.  This  line  is  the  //-projection  of  the 
circular  path  of  the  revolving  point,  the  center  of  revolution 
being  the  point  e  (eh)  lying  on  HQ.     The  line  aheh  is  one  pro- 


Fig.  217. 


Fig.  218. 


jection  of  a  line  lying  in  Q  (a  line  of  maximum  inclination  to 
H,  §  11-4).  Find  avev  (Prob.  10,  §  loo),  and  the  true  length, 
aveu  of  the  line  ae  (Prob.  3,  §  78).  The  distance  cVe^  is  the 
actual  distance  of  the  point  a  from  the  center  of  revolution  eh. 
Lay  off  ehar  =  eCe^,  thus  giving  the  required  revolved  position  nr. 

Note.  The  revolved  position  aT  represents  the  revolution  of  Q  about 
HQ  through  an  ancjle  greater  than  90°.  The  distance  avei  may  be  laid  off 
from  eh  in  the  opposite  direction  if  preferred  (that  is,  on  the  same  side  of 
HQ  as  the  projection  a*),  representing  the  revolution  of  Q  through  an 
angle  less  than  90°.  It  is  customary  to  place  the  revolved  position  where 
it  will  be  as  far  from  the  other  parts  of  the  figure  as  possible. 

In  Fig.  218,  the  point  a,  lying  in  Q,  is  revolved  about  VQ 
into  V.  The  construction  is  entirely  analogous  to  that  for 
revolving  about  HQ  into  H,  and  is  left  without  further  explana- 
tion to  the  student. 

Working  Rule.  Returning  to  Fig.  217,  the  revolved  posi- 
tion, aT,  is  obtained  by  laying  off  from  eh,  on  the  perpendicular 


XV,  §  138]      THE  REVOLUTION  OF  PLANES 


141 


akeh,  a  certain  distance,  namely,  the  true  length  of  the  line  ae. 
The  situation  of  this  line  is  such  that  it  is  not  necessary  to 
draw  the  complete  construction,  or  even  the  F-projection,  avev, 
in  order  to  find  the  true  length.  For,  it  will  be  seen  that  the 
true  length,  aveu  is  the  hypothenuse  of  a  right  triangle,  of 
which  the  base  is  euf=  aheh,  and  the  altitude  avf.  We  may 
therefore  state  the  following  working  rule  :  To  revolve  point  a, 
lying  in  the  plane  Q,  about  HQ  into  H,  proceed  as  follows. 
From  ah  draw  a  line  perpendicular  to  HQ.  Then  the  revolved 
position,  aT,  lies  on  this  perpendicular,  at  a  distance  from  HQ 
equal  to  the  hypothenuse  of  a  right  triangle,  of  which  the  two 
sides  are  the  distances  ah  to  HQ,  and  o*  to  GL.  This  hypoth- 
enuse can  be  found  directly  with  the  dividers,  as  follows  :  Take 
the  distance  aheh ;  set  it  off  from  /  to  point  ex  on  the  ground 
line  ;  span  the  distance  from  ex  to  av. 

Similarly,  to  revolve  a  point  lying  in  a  plane  about  the  V- 
trace  of  the  plane  into  V  (Fig.  218),  find  the  hypothenuse  of  a 
right  triangle,  of  which  the  two  sides  are  the  distance  from  the 
F-projection  of  the  point  to  the  F"-trace  of  the  plane,  and  from 
the  if-projectlon  of  the  point  to  the  ground  line.  This  hy- 
pothenuse is  the  true  distance  of  the  revolved  point  from  the 
F"-trace  of  the  plane. 

Special  Case.     The   trace  of   the   plane  about   which  the 
revolution  is  made  is  perpendicular 
to  the  ground  line.     Let  HQ,  Fig. 
219,  be  perpendicular  to  GL,   VQ 
being  at  any  angle.     The  plane  Q         / 
is  therefore  perpendicular  to  V,  so 
that  the    revolution    of   the   given 
point   occurs    in  a  plane  which  is 
parallel   to   V.     The    path   of   the 
revolution  must  then  project  on  V 
in   its   true    shape,    that  is,   as    a 
circular  arc,  and  the  revolved  posi- 
tion, ar,  can  be  obtained  at  once 
by    projecting   from   the   intersection   of 
ground  line. 


/ 


/ 


Fig.  219. 
this    arc    and 


the 


142  DESCRIPTIVE   GEOMETRY  [XV,  §  r38 

The  ease  with  which  the  revolved  position  is  obtained  in 
this  particular  situation  of  the  plane  suggests  another  method 
for  solving  the  general  case,  since  an  auxiliary  plane  of  pro- 
jection can  always  be  assumed  perpendicular  to  either  trace  of 


Fig.  '220. 

any  given  plane  (see  §  70).  Such  a  construction  is  shown  in 
Fig.  220,  and  is  left  without  further  explanation  to  the  student. 

Corollary.  To  find  the  revolved  position  of  a  line  lying  in  a 
plane,  when  the  plane  is  revolved  into  H  or  V  about  the  corre- 
sponding trace. 

Since  a  straight  line  is  determined  when  two  of  its  points 
are  known,  this  problem  can  always  be  solved  by  revolving 
two  points  of  the  line  by  the  method  already  given.  However, 
a  simpler  method  can  usually  be  found,  as  in  the  following 
examples. 

(1)  Let  the  line  A,  Fig.  221,  which  is  to  be  revolved  about 
VQ  into  V,  intersect  VQ  in  the  point  t.  Revolve  any  point  of 
A,  as  c(ch,  cv)  to  cr.  If  now  we  attempt  to  revolve  the  point  t, 
this  point,  being  on  the  axis  of  revolution,  will  not  move. 
Hence  the  revolved  position  of  the  line,  Ar,  passes  through  cr 
and  t. 

(2)  Let  the  line  A,  Fig.  222,  which  is  to  be  revolved  about 
VQ  into  V,  be  parallel  to  VQ.  Revolve  any  point  c,  of  A,  to 
cr.     Every  other  point  of  A  is  the  same  distance  from  the  axis 


XV,  §  140]         PROBLEMS  OF  REVOLUTION 


143 


VQ,  and  will  revolve  just  as  far.     Hence,  the  revolved  position 
Ar  passes  through  cr  and  is  parallel  to  VQ. 


Fig.  221. 


Fig.  222. 


139.  Solution  of  Plane  Problems  by  Revolution  of  the  Plane. 

By  extending  the  preceding  problem,  any  number  of  points  or 
lines  lying  in  a  plane  can  be  revolved  into  one  of  the  coordinate 
planes,  and  the  revolved  position  will  show  the  true  relation 
existing  between  them.  The  problem  thus  becomes  a  very 
useful  auxiliary  in  the  solution  of  a  certain  class  of  problems. 

In  revolving  about  either  trace  of  a  plane,  as  for  example 
about  HQ  into  H,  it  should  be  noticed  that,  while  the  F-pro- 
jections  of  the  points  and  lines  concerned  must  be  considered, 
no  use  is  made  of  the  U-trace  of  the  plane,  VQ.  Similarly, 
when  revolving  about  VQ  into  V,  no  use  is  made  of  HQ. 
Hence,  when  a  plane  is  introduced  as  an  auxiliary,  for  the  ex- 
press purpose  of  revolving  about  one  of  its  traces,  it  is  usually 
unnecessary  to  find  more  than  one  trace  of  the  plane. 

140.  The  Angle  between  Two  Intersecting  Lines.  The  angle 
between  two  intersecting  straight  lines  is  a  plane  angle.  It  is 
measured  in  the  plane  that  is  determined  by  the  two  lines. 

Problem  22.     To  find  the  angle  between  two  intersecting  lines. 

Analysis.  Find  the  plane  containing  the  given  lines.  About 
one  trace  of  this  plane,  revolve  the  lines  into  the  corresponding 
coordinate  plane,  when  the  true  angle  between  them  will  appear. 


144 


DESCRIPTIVE  GEOMETRY 


[XV,  §  140 


Construction  (Fig.  223).  Let  A  and  B  be  the  given  lines. 
Find  either  trace  of  the  plane  containing  these  lines,  for 
example  the  IT-trace,  HX  (Prob.  6,  §  106).     Revolve  both  lines 


Fig.  223. 


Fig.  224. 


about  HX  into  H.  To  accomplish  this  most  readily,  revolve 
their  point  of  intersection,  o,  to  or  (Prob.  21,  Working  Rule, 
§  138).  Then  AT  passes  through  or  and  the  trace  s1}  while  Br 
passes  through  or  and  the  trace  s2  (Corollary,  §  138).  The 
angle,  8,  between  Ar  and  BT  is  the  true  angle  between  the  lines. 

A  similar  construction  is  shown  in  Fig.  224,  in  which  the 
P-trace,  VX,  of  the  plane  containing  A  and  B  is  used.  The 
angle,  8,  between  Ar  and  Br,  is  the  true  angle  between  the  lines. 

Special  Case.  One  line  is  parallel  to  H,  and  the  other  is 
parallel  to  V.  The  angle  may  be  found  by  the  general  method, 
but  the  following  is  simpler. 

Analysis.  Revolve  either  line  about  the  other  as  axis,  until 
both  are  parallel  to  the  same  coordinate  plane. 

Construction  (Fig.  225).  Let  A  and  B  be  the  given  lines. 
Choose  one  of  them,  as  A,  for  the  axis.  Since  A  is  parallel  to 
H,  B  must  now  be  revolved  about  A  until  B,  also,  is  parallel  to 
H.  Assume  any  point  c(ch,  cv)  in  the  line  B,  at  a  reasonable 
distance  from  the  intersection,  o,  of  the  lines.  Through  c* 
draw  an  indefinite  line  perpendicular  to  Ah.  Take  the  distance 
ovcv  in  the  compasses,  and  with  oh  as  center  strike  an  arc  across 


XV,  §  140]         PROBLEMS  OF  REVOLUTION 


145 


Fig.  225. 


this  perpendicular,  giving  cr.  Then  ohcr  is  B  revolved,  Br, 
and  the  angle  between  Br  and  Ah  is  the  true  angle  8,  between 
the  given  lines. 

Proof  of  the  Construction.  Since  A 
is  parallel  to  H,  the  revolution  of 
point  c  about  A  will  project  on  H  as  a 
straight  line  perpendicular  to  Ah  (§  76). 
Hence  c  revolved,  cr,  must  lie  some- 
where on  the  perpendicular  to  Ah 
through  c\  Also,  when  the  portion  oc 
of  the  line  B  is  revolved  until  parallel 
to  H,  it  will  appear  in  true  length. 
Since  the  point  o  in  the  axis  A  does 
not  move,  ohcr  should  equal  the  true 
length  of  oc.  But  ohcr  was  made  equal 
to  ovcv,  which  is  the  true  length  of  oc, 
since  the  line  B  is  parallel  to  V. 
Hence  the  construction  is  correct. 

Corollary.  To  find  the  projections  of  the  bisector  of  the  angle 
between  two  intersecting  lines. 

Let  it  be  required  to  find  the  bisector  of  the  acute  angle.  In 
Fig.  223,  draw  first  the  actual  bisector,  0„  in  the  revolved 
position,  bisecting  the  angle  between  Ar  and  Br.  Xow  CT 
necessarily  passes  through  or,  and  in  this  case  intersects  the 
trace  IIX  in  the  points/.  Imagine  the  plane  Xto  be  counter- 
revolved  (revolved  back)  about  IIX  to  its  original  position, 
taking  with  it  the  lines  A,  B,  and  C.  The  lines  A  and  B,  and 
their  intersection,  point  o,  will  resume  their  former  positions. 
The  point  83k,  being  on  the  axis  of  revolution  HX,  will  not 
move.  Hence  the  //-projection,  C*  of  the  line  C  must  pass 
through  o*  and  .%h. 

To  find  the  ^projection  of  C,  note  thato*  must  be  projected 
to  o",  while  N,\  being  an  ff-trace,  lies  in  H,  and  musl  project 
to  Sj"  in  GL.     II'  ace  <'   \    determined  bj  "  and  -V- 

The  second  example,  Fig.  224,  cannot  be  solved,  however,  in 
this  way.  As  soon  as  the  bisector,  Cr,  i  drawn  in  the  revolved 
position,  it  is  seen  that  the  intersection  of  0,  with  the  trace  I'.V 


146 


DESCRIPTIVE  GEOMETRY 


[XV,  §  140 


is  beyond  the  limits  of  the  figure.     One  point  of  C"  is  o",  and 

one  point  of  Ch  is  oh.     It  remains  to  find  a  second  point  in  each 

projection.     Assume  a  point,  er,  in  Cr.     Through  er  draw  the 

revolved  position,  Lr,  of  a  line  lying  in  the  plane  of  the  given 

lines,   by   making   LT  parallel   to 

the   revolved    position   of    either 

of  the  given  lines  A  or  B ;  in  this 

case  LT  is  drawn  parallel  to  AT. 

Since  the  line  L  is  in  the  plane  X, 

the  intersection,  t3v,  of  Lr  and  VX 

is  the  F-trace  of  the  line  L,  and 

t3h  is  in  GL.     Also,  since  the  line  L 

is  parallel  to  the  line  A,  through 

tzv  draw  Lv  parallel   to  A1',   and 

through  t3h  draw  Lh  parallel  to  Ah. 

Now  e  is  a  point  in  the  line  L. 

Hence  revolve  back  from  er  per-  FlG-  22i  Orated). 

pendicular  to    VX,  and  find  ev  on  L" ;    jjroject  from  e"  and 

obtain  eh  on  Lh.     Since  e  is  a  point  also  in  the  line  C,  we  have 

Cv  determined  by  ov  and  e",  and  Ch  determined  by  oh  and  e\ 

141.  The  Angle  between  a  Line  and  a  Plane.  The  angle 
which  a  line  makes  with  a  plane  is  equal  to  the  angle  between 
the  line  and  its  projection  on  the  plane.  Hence  we  only  need 
to  find  the  angle  between  two  intersecting  straight  lines. 

Problem  23.     To  find  the  angle  between  a  line  and  a  plane. 

First  Analysis.  Project  the  given  line  on  the  given  plane. 
Find  the  true  angle  between  the  given  line  and  its  projection ; 
this  is  the  angle  required. 

Second  Analysis.  From  any  point  of  the  given  line,  drop  a 
perpendicular  to  the  given  plane.  Find  the  true  angle  between 
the  given  line  and  the  perpendicular  ;  this  is  the  complement  of 
the  required  angle. 

In  general,  the  second  analysis  gives  the  simpler  construc- 
tion, and  is  the  one  usually  adopted. 

Construction  (Fig.  226).  Let  A  be  the  given  line  and  Q  the 
given  plane.     Assume  any  point  e  (c\  c")  on  A.     From  c  draw 


XV,  §  141]        PROBLEMS  OF  REVOLUTION 


147 


the  line  B  (Bh,  Bv)  perpendicular  to  the  plane  Q  (§  111).  Find 
the  true  angle,  S,  between  the  lines  A  and  B  (Prob.  22,  §  140). 
Then  y,  which  is  the  complement  of  S,  is  equal  to  the  true 
angle  between  line  A  and  plane  Q. 


Fig.  226. 


Fig.  227. 


In  Fig.  227,  the  given  plane  Q  is  parallel  to  the  ground  line. 
The  general  case  still  applies,  but  there  is  an  extra  detail  in 
the  construction.  As  before,  assume  any  point  c  in  the  line  A. 
From  c  drop  the  perpendicular  B  to  the  plane  Q.  Find  the 
angle,  S,  between  the  lines  A  and  B.  Then  y,  the  complement 
of  8,  is  the  angle  required.  As  a  necessary  step  in  finding  the 
angle  S,  we  need  one  trace,  as  VX,  of  the  plane  containing  the 
lines  A  and  B.  To  determine  VX  we  need  the  F-trace  of  each 
of  these  lines.  The  F"-trace  of  A  is  readily  found  (Prob.  1, 
§37).  But  B  is  a  profile  line,  of  which  hut  one  point,  c,  is 
known.  Nevertheless,  the  line  B  is  a  definite  line,  since  it  is 
perpendicular  to  Q.  To  find  the  F-trace  of  B,  proceed  as  fol- 
lows:  Find  the  profile  projection,  c,p,  of  c;  also  the  profile 
trace,  PQ,  of  Q.  From  <•"  draw  the  profile  projection  of  B,  Bp 
perpendicular  to  PQ.  Then  the  F-trace  of  B  is  the  point  t2, 
in  which  B"  intersects  VP.    (See  Prob.  11,  Special  Case  2,  §  115.) 


148 


DESCRIPTIVE  GEOMETRY 


[XV,  §  142 


142.  A  Rectilinear  Figure  Lying  in  a  Plane.  A  plane  poly- 
gon of  more  than  three  sides  cannot  be  assumed  by  assuming 
at  random  the  projections  of  a  number  of  points  in  space,  and 
then  connecting  these  points  by  straight  lines.  For,  as  soon  as 
three  points  not  in  the  same  straight  line  have  been  assumed, 
a  plane  is  determined ;  thereafter  some  consideration  is  neces- 
sary to  assure  us  that  the  remaining  points  lie  in  the  plane 
determined  by  these  first  three  points. 

Problem  24.  Given  one  complete  projection,  and  three  points  in 
the  other  projection  of  a  plane  polygon,  to  complete  the  projection. 

First  Analysis.  Lines  which  intersect,  or  are  parallel,  are 
necessarily  in  a  plane.  Hence,  by  producing  the  sides  of  the 
given  complete  projection  until  they  intersect,  or  by  drawing 
in  this  projection  lines  which  are  parallel  to  some  of  its  sides  or 
which  intersect  the  sides  or  each  other,  auxiliary  lines  are  ob- 
tained by  means  of  which  the  second  projection  can  be  built  up. 

In  particular,  if  the  figure  is  a  quadrilateral,  the  intersection 
of  its  diagonals  is  a  point  in  the  plane  of  the  figure. 


Fig.  228. 


Fig.  229. 


Second  Analysis.  Both  projections  of  three  points  are 
known.  Find  the  traces  of  the  plane  containing  these  points. 
For  the  remaining  points,  one  projection  of  each  point,  and 
the  traces  of  the  plane  containing  it,  are  known ;  hence  the 
second  projection  can  be  found. 


XV.  I  142]        PROBLEMS  OF  REVOLUTION 


149 


Construction  by  First  Analysis  (Fig.  228).  Let  the  ^-projec- 
tion. akbkckdk,  and  arbTcT  of  the  U-projeetion  be  given.  Since 
this  figure  is  a  quadrilateral,  we  may  draw  the  diagonals  ahck 
and  bhdk,  intersecting  in  the  point  e\  In  the  F-projection 
draw  arcr.  Locate  eT  on  this  line  by  projecting  from  e*.  Draw 
brer,  produce  it,  and  locate  on  it  dT  by  projecting  from  dk. 

Otherwise  (Fig.  229),  through  dh  draw  dkfk  parallel  to  bkck. 
and  intersecting  ahbh  at  /*.  Project  from  /*  to  /r  in  arbT. 
Through  fT  draw  fTdT  parallel  to  bTcT.  Then  the  line  dt\  since 
it  passes  through  the  point  /in  ah,  and  is -parallel  to  be.  lies  in 
the  plane  of  the  quadrilateral.     Locate  dz  by  projecting  from  dh, 

A  second  example  is  given  in  Fig.  230.  Let  the  complete 
F-projection,  aTbTcTdTeT.  and  the  17-projection  of  three  points, 


a*,  &*,  and  c\  be  given.     In  this  poh  -  and  dTeT  can  both 

be  produced  to  meet  <VhT  produced  in  the  accessible  por 
and  A-r  respectively.     Produce  «*&*,  and  project./*  and  kr  on  it. 
Draw/c*.     This  line  produced  contains  dk.  which  is  found  by 
projection  from  d'.     The  point  dk  being  thus  .raw  dkkk, 

and  proj<    I  m  e*.     Draw  •'  '.  ting  th<  gon. 

•   These  constructions  are  given  mei  s  ..lustrations  of  the 


150 


DESCRIPTIVE  GEOMETRY 


[XV,  §  142 


method.  Other  constructions  should  suggest  themselves  to 
the  student.  There  is  often  a  chance  for  the  exercise  of  con- 
siderable ingenuity  in  solving  problems  in  this  manner,  espe- 
cially in  the  matter  of  obtaining  accuracy  in  the  result.  Note 
that  the  ground  lme  is  not  essential  in  these  constructions. 

Construction  by  Second  Analysis  (Fig.  231).  Let  the  complete 
F-p  rejection,  avbvcvdvev,  and  three  points,  ah,  bh,  and  c*,  of  the 
iT-projection  be  given.  By  producing  the  lines  ab  and  be, 
obtain  the  plane  X,  containing  these  two  intersecting  lines 
(Prob.  6,  §  106).  Points  d  and  e  lie  in  this  plane,  and  d"  and 
and  ev  are  known ;  find  dh  and  eh  (Prob.  16,  §  133).  Connect 
chdh,  dheh,  ehah. 


Fig.  231. 


Corollary  1.     To  find  the  true  size  and  shape  of  a  plane 
polygon. 

First  Analysis.     Find  one  trace  of  the  plane  containing  the 


XV,  §  142]        PROBLEMS  OF  REVOLUTION  151 

polygon.  Revolve  the  polygon  about  this  trace  into  the  corre- 
sponding coordinate  plane. 

Second  Analysis.  If  any  side  of  the  polygon  is  parallel  to 
one  of  the  coordinate  planes,  the  polygon  may  be  revolved 
about  this  line  as  an  axis  into  a  position  parallel  to  this  coor- 
dinate plane. 

Construction  by  First  Analysis  (Fig.  231).  Each  corner  of 
the  polygon  is  here  revolved  about  VX  into  V  (Prob.  21, 
Working  Rule,  §  138),  due  regard  being  had  to  the  relative 
positions  of  the  points  with  respect  to  the  axis  VX. 

Check.  Note  that,  for  each  side  of  the  polygon,  the  P~-pro- 
jection  and  the  revolved  position  (produced  if  necessary) 
must  intersect  VX  at  the  same  point,  namely,  the  P~-trace  of 
this  line. 

Construction  by  Second  Analysis  (Fig.  230).  In  this  polygon, 
the  line  ab  is  parallel  to  H.  Hence  the  polygon  may  be  re- 
volved about  ahbh  into  a  position  parallel  to  //.  Revolve  first 
the  point  d.  To  do  this,  through  dh  draw  a  line  perpendicular 
to  ahbh.  Obtain  the  true  length  of  the  line  dj  (Prob.  3,  §  78). 
In  this  case  dvjv  is  the  true  length,  since  dj  is  parallel  to  V 
With/1  as  center,  radius  equal  to  the  true  length  of  dj,  strike 
an  arc  across  the  perpendicular  from  dh,  thus  obtaining  dr. 
From  dr  draw  to  the  fixed  points  jK  and  kh  on  the  axis  ahbh. 
Obtain  cr  on  djh,  and  er  on  drkh,  by  drawing  from  ch  and  eK  per- 
pendicular to  ahbk.  Then  since  ah  and  bh  are  fixed  points,  the 
true  size  of  the  polygon  is  ahbhcrdrer. 

Corollarv  2.  To  find  the  projection  of  the  line  which  bisects 
one  of  the  interior  angles  of  a  plane  polygon. 

Construction.  In  Fig.  2.'U  the  interior  angle  at  e  is  bisected. 
The  bisector  is  first  drawn  in  the  revolved  position,  and  inter- 
sects the  side  b,cr  in  the  point  fr.  Project  from  fr  perpen- 
dicular to  VX  to  f°  in  b"c".  and  from  /'"  perpendicular  to  GL 
to/*  in  6*c* ;  then  evf'  and  ehfh  are  the  projections  required. 

In  Fig.  230,  the  interior  angle  at  h  is  bisected,  the  actual 
bisector,  bhf,  intersecting  drer  in  the  point  f.  From  fr  are 
obtained/*  and  /",  giving  bhfh  and  b"fv  as  the  required  pro- 
jections. 


CHAPTER   XVI 


MISCELLANEOUS   PROBLEMS    ON  THE   LINE   AND   PLANE 


143.  The  Distance  from  a  Point  to  a  Line.  The  shortest  dis- 
tance from  a  point  to  a  given  straight  line  is  obtained  by 
dropping  a  perpendicular  from  the  point  to  the  line.  This 
perpendicular  evidently  lies  in  the  plane  which  is  determined 
by  the  given  point  and  line. 

Problem  25.  To  find  the  shortest  (perpendicular)  distance  from 
a  point  to  a  line. 

Analysis.  In  the  general  position  of  the  line,  the  right  angle 
between  it  and  the  perpendicular  will  not  project  as  such 
(§  109).  Hence,  pass  an  auxiliary  plane  through  the  given 
line  and   point.     Kevolve   this  plane  into   either   coordinate 


Fig.  232. 


plane,  obtaining  thus  the  true  relative  position  of  the  line  and 
point.  Draw  the  required  perpendicular  in  the  revolved  posi- 
tion. To  find  the  projections  of  the  perpendicular,  revolve 
the  auxiliary  plane  back  to  its  original  position. 

152 


XVI,  §  143]    PROBLEMS  OX  THE  LIXE  AXD  PLAXE    153 

Construction  (Fig.  232).  Let  A  be  the  given  line,  and  c  the 
given  point.  Through  c  draw  the  auxiliary  line  B  {Bh,  Bv) 
parallel  to  A  (§  93).  Pass  the  auxiliary  plane  X  through  the 
lines  A  and  B  (Prob.  6,  §  106).  Since  X  is  introduced  solely 
for  the  purpose  of  obtaining  a  revolved  position,  but  one  trace, 
as  HX,  is  necessary  (§  139).  Revolve  A  and  c  about  HX  into 
H.  To  do  this  most  readily,  revolve  point  c  to  cr  (Trob.  21, 
Working  Rule,  §  138)  ;  draw  Br  through  c,  and  the  trace  s2 
(Prob.  21,  Corollary,  §  138),  then  draw  Ar  through  the  trace  sx 
and  parallel  to  Br.  From  cr  drop  the  perpendicular  cTdr  to  Ar ; 
this  is  the  actual  shortest  distance  required.  To  find  the  pro- 
jections of  the  perpendicular  cd,  revolve  back  from  dr  perpen- 
dicular to  HX,  and  find  dh  in  Ah ;  draw  chdh.  Project  from  dh 
to  d"  in  Av,  and  draw  cvd". 

Special  Case  I.  The  given  line  is  parallel  to  one  of  the 
coordinate  planes  (Fig.  233).  Let  the  given  line  A  be  parallel 
to  H,  and  let  c  be  the  given  pomt.    It  is  required  to  draw  from 


dv 
Fig.  23.3. 

c  a  line  which  shall  be  perpendicular  to  the  given  line  A.  Xow 
if  two  lines  are  at  right  angles,  the  right  angle  will  project  as 
such  if  one  of  the  lines  is  parallel  to  a  coordinate  plane  (**  110). 
Since  A  is  parallel  to  H,  draw  chdh  perpendicular  to  J\  giving 
at  once  the  H"-projection  of  the  perpendicular  from  c  to  .1. 
Project  from  d*  to  d"  in  Av ;  draw  c?dv.  For  the  actual  dis- 
tance from  the  point  to  the  line,  find  the  true  Length  of  <•</ 
(Prob.  3,  §80). 


154 


DESCRIPTIVE   GEOMETRY         [XVI,  §  143 


The  above  solution  suggests  a  method  for  solving  the  general 
case,  since  any  line  can  be  converted  into  a  ^-parallel  by  as- 
suming a  secondary  plane  of  projection  parallel  to  the  ^-pro- 
jection of  the  line  (§  70).  This  is  illustrated  in  Fig.  234. 
In  that  figure,  ab  is  the  given  line,  c  the  given  point,  and  a 
secondary  projection  is  made  after  assuming  G^  parallel  to 


Fig.  234. 


Fig.  235. 


ahbh.  Further  explanation  of  the  construction  is  omitted, 
since,  for  the  line  in  question,  the  solution  is  of  doubtful  ad- 
vantage as  compared  with  the  general  solution. 

If,  however,  the  given  line  is  a  profile  line,  the  simplest  solu- 
tion is  obtained  by  the  use  of  an  additional  projection. 

Special  Case  II.  The  given  line  lies  in  a  profile  plane. 
Let  ab,  Fig.  235,  be  the  given  line,  and  c  the  given  point.  Find 
the  profile  projection,  apbp,  of  the  line  ab.  Project  c  on  to  P, 
and  find  the  profile  projection  cp.  From  cp  draw,  perpendicu- 
lar to  apbp,  the  profile  projection,  cpdp,  of  the  required  perpen- 
dicular. From  eZpfind  the  projections  d"  and  dh ;  then  c*dA  and 
cvdv  are  the  required  projections.  The  true  length  of  cd  (Prob. 
3,  §  80)  is  the  required  shortest  distance. 


XVI,  §  144]    PROBLEMS  ON  THE  LINE  AND  PLANE     155 

144.  Distance  between  Two  Lines.  If  two  lines  are  not  in  the 
same  plane,  that  is,  neither  intersect  nor  are  parallel,  there  is 
but  one  line  that  is  perpendicular  to  both  of  them.  This  line, 
the  common  perpendicular,  measures  the  shortest  distance  be- 
tween the  two  given  lines. 

Problem  26.  To  find  the  shortest  distance  between  two  lines  not 
in  the  same  plane,  and  the  projections  of  their  common  perpendicular. 

Analysis  (Fig.  236).  Let  A  and  B  be  the  given  lines, 
in  any  oblique  position  in  space.  Through  either  given  line, 
as  A,  pass  an  auxiliary  plane  Q,  parallel  to  the  other  given 
line  B.     Then  B  is  everywhere  equally  distant  from  Q,  and 


Fig.  236. 

this  distance  is  equal  to  the  shortest  distance  between  B 
and  A.  Assume  any  point,  e,  in  B;  from  e  drop  a  perpen- 
dicular to  the  plane  Q,  intersecting  Q  at  the  point  k.  Note 
that  although  this  lirst  perpendicular  ek  is  not  in  the  required 
position,  it  does  have  the  required  direction.  Now  let  the 
perpendicular  ek  be  moved,  parallel  to  itself,  so  that  the 
point  e  moves  in  the  line  B.  The  point  k  will  move  in  the 
plane  Q,  describing  in  this  plane  a  line  L,  parallel  to  B. 
Since  A  is  also  in  Q,  and  since  L  is  not  parallel  to  A  because 
B  is  not,  L  and  A  will  intersect  in  a  point,  n.  Let  the  per- 
pendicular ek  come  to  rest  when  point  k  coincides  with  point  n. 
The  line  will  then  be  in  the  position  on,  intersecting  both  A 
and  B  and  perpendicular  to  each.  Hence  on  is  the  common 
perpendicular  required.  The  actual  distance  between  the 
lines  A  mid  Ji  may  ho  found  by  measuring  the  length  of 
either  ek  or  on. 


156 


DESCRIPTIVE  GEOMETRY         [XVI,  §  144 


Construction  (Fig.  237).  Let  A  and  B  be  the  given  lines. 
Through  A  pass  the  plane  Q,  parallel  to  B  (Prob.  7,  §  107). 
To  do  this,  assume  any  point,  c,  in  A.  Through  c  draw  line  D 
parallel  to  B.  Pass  the  plane  Q  through  the  lines  A  and  D. 
Next  assume  any  point  e  in  B.  From  e  draw  the  line  F  per- 
pendicular to  Q  (§  111).  Find  the  point  k,  in  which  F  inter- 
sects Q  (Prob.  13,  §  119).  Here  the  auxiliary  plane  X  perpen- 
dicular to  H  is  used.  Planes  X  and  Q  intersect  in  the  line  J"; 
Jv  and  Fv  intersect  in  kv,  one  projection  of  the  required  point  k. 


Fig.  237. 

Let  now  the  trial  perpendicular  ek  be  carried  parallel  to  itself 
to  the  position  on.  In  this  translation,  point  e  moves  along  the 
line  B,  and  point  k  moves  in  the  line  L,  parallel  to  B,  to  the 
point  n,  where  L  intersects  A.  Consider  each  projection  sepa- 
rately. In  the  //-projection,  through  kh  draw  Lh  parallel  to  Bh ; 
Lh  intersects  Ah  at  nh ;  through  n*  draw  nhoh  parallel  to  ehkh  to 
intersect  Bh  at  ok.  Make  the  analogous  construction  in  the 
F-projection.     Thus  nv  and  o"  are  determined  independently 


XVI,  §  144]    PROBLEMS  ON  THE  LINE  AND  PLANE    157 


of  nh  and  o\  But  nh  and  ri°  are  two  projections  of  the  same  point ; 
likewise  oh  and  o".  Hence,  as  a  check  on  the  construction,  nh 
and  nv  should  lie  in  the  same  projector,  as  should  also  oh  and  o". 

The  line  no  (nhoh,  uvov)  is  the  required  common  perpendicular. 
The  shortest  distance  between  the  given  lines  is  equal  to  the 
true  length  of  no  (Prob.  3,  §  80). 

Special  Case  I.  One  of  the  given  lines  is  parallel  to  H  or 
V.  The  general  solution  will  apply  to  this  case  ;  but  a  partic- 
ular solution,  much  simpler  than  the  general  one,  can  be  had  by 
using  a  secondary  plane  of  projection. 

Let  A  and  B,  Fig.  238,  be  the  given  lines,  the  line  A  being 
parallel  to  H.     Assume  a  secondary  ground  line  GXLU  perpen- 


Fig.  238. 


dicular  to  Ah.  Project  the  given  lines  A  and  B  on  the  second- 
ary vertical  plane  Vx  thus  assumed  (§  68).  The  line  A  will 
project  as  a  point,  Axv.  The  required  line  no  is  perpendicular 
to  both  A  and  B.  Since  A  is  perpendicular  to  Vx,  no,  which  is 
perpendicular  to  A,  is  parallel  to  Vx.  Again,  since  no  and  5 
are  perpendicular,  and  no  is  parallel  to  Vx,  the  projection  nfof 
and  Bf  are  perpendicular  (§  110).  Hence,  from  Af  draw 
ni'Ox"  perpendicular  to  By.  Project  from  of  on  Bf  to  o*  on 
Bh,  thence  to  ov  on  B".  From  oh  draw  ohnh  perpendicular  to  Ah 
(see  Prob.  25,  Special  Case  I,  §  143).  Project  from  »*  to  nv  in 
^l",  and  draw  ovnv.  Then  n*o*  and  n'o*  are  the  projections  of 
the  common  perpendicular,  while  the  actual  shortest  distance 
is  equal  to  nfof. 


158 


DESCRIPTIVE   GEOMETRY         [XVI,  §  144 


Special  Case  II  (Eig.  239).  The  given  lines,  ab  and  cd, 
are  both  profile  lines.     Project  both  lines  on  to  the  same  profile 

plane ;  the  plane  here  used  is  the 
one  containing  the  given  line  ab. 
Find  the  profile  projections,  apbp 
and  cpdp.  Then  it  is  evident  that 
the  common  perpendicular,  ef,  will 
project  on  P  as  a  point,  located  at 
the  intersection  of  apbp  and  cpdp. 
The  rest  should  be  obvious. 

Note.  If  one  of  the  given  lines  is  a 
profile  line,  the  other  being  a  general 
oblique  line,  the  general  solution  applies. 
The  simpler  construction  results  when 
the  auxiliary  plane  Q  is  passed  through 
the  profile  line. 

145.  The  Angle  between  Two  Planes.  Let  two  planes  inter- 
sect so  as  to  form  a  dihedral  angle.  The  figure  is  left  for  the 
student  to  draw.  The  angle  between  these  planes  is  measured 
by  two  lines,  one  in  each  plane,  each  perpendicular  at  the  same 
point  to  the  line  of  intersection  of  the  given  planes.  It  is 
evident  that  these  two  lines  lie  in  a  third  plane  which  is  per- 
pendicular to  both  of  the  given  planes  and  to  their  line  of 
intersection.  They  are,  in  fact,  the  lines  of  intersection  of 
this  third  plane  with  the  given  planes. 

Problem  27.     To  find  the  angle  between  two  planes. 

First  Analysis.     From   any  point  in  space,  drop  a  perpen- 
dicular to  each  of  the  given  planes. 
The  angle  between*  these  lines  is 
ecpial  to  the   angle   between   the 
planes. 

Second  Analysis.  Let  Q  and  R, 
Fig.  240,  be  the  given  planes,  and 
let  these  planes  intersect  in  the 
line  A.  At  any  point  c  in  A, 
pass  an  auxiliary  plane  Z,  perpendicular  to  A.  The  plane  Z 
intersects  Q  in  the  line  cs2  and  R  in  the  line  cs3.     The  angle 


Fig.  240. 


XVI,  §145]    PROBLEMS  ON  THE  LINE  AND  PLAXE    159 


between  cs2  and  cs3  is  the  required  angle  between  Q  and  R. 
Let  Ah  be  the  projection  of  A  on  H.  Since  Z  is  perpen- 
dicular to  A,  Ah  and  .HZ  are  perpendicular  (§  111).  The 
projection  Ah  and  the  trace  HZ  intersect  in  the  point  o  ;  con- 
nect c  and  o.  Now  co  is  perpendicular  to  A,  since  it  lies  in 
the  plane  Z  which  is  perpendicular  to  A.  Also,  co  is  perpen- 
dicular to  HZ,  since  it  lies  in  the  plane  of  the  lines  A  and  Ak, 
which  is  perpendicular  to  HZ.  Revolve  the  plane  Z  about 
HZ  into  H.  Since  co  is  perpendicular  to  HZ,  c  will  revolve 
with  o  as  a  center,  and  will  fall  on  Ah,  because  the  latter  is 
also  perpendicular  to  HZ,  at  the  distance  cp  =  co.  The  lines 
cs.z  and  cs3  will  take  the  revolved  positions  crs2  and  crss,  and  will 
show  the  true  size  of  the  angle  6,  between  thein. 

In  actual  construction  there  is  little  choice  between  the 
first  and  the  second  analysis.  Moreover,  the  latter  usually 
adapts  itself  better  to  the  way  in  which  the  problem  most 
often  occurs  in  practice.  Hence  the  constructions  here  given 
will  be  made  in  accordance  with  the  second  analysis. 

Construction  (Fig.  241).  Let  Q  and  R  be  the  given  planes. 
Find  the  line  of  intersection,  A,  of  these  planes  (Prob.  12 
§  118).  Assume  an  auxiliary 
plane,  Z,  perpendicular  to  A, 
by  drawing  HZ  perpendicular 
to  Ah.  The  F-trace,  VZ,  is 
not  needed  (§  139).  The  trace 
HZ  intersects  Ak  in  the  point 
o,  lying  in  H.  It  is  necessa  rv 
to  find  the  shortest  distance 
from  this  point  to  the  line  A. 
Assume  a  secondary  ground 
line,  G]LX,  coincident  with  Ah. 
Project  the  line  A  as  A? 
(§  68),  using  the  traces  s  an<  1  /  FlG-  241- 

as  the  most  convenient  points  in  the  line.  Point  o  remains  in 
G\LX.  From  o  draw  oc,'  perpendicular  to  .  I,' ;  this  is  the  slim  test 
distance  from  o  to  A.  On  Ah,  lay  off  ocr  =  octv,  thus  obtaining  <•,. 
Note  the  points  s.,  and  gg,  in  which  HZ  intersects  HQ  and  Hit 


160 


DESCRIPTIVE   GEOMETRY         [XVI,  §  145 


respectively.  Draw  crs2  and  crs3.  The  angle  between  these 
lines  equals  the  required  angle,  6,  between  the  given  planes. 

Additional  examples  are  given  in  Figs.  242  and  243.  Each 
of  them  falls  under  the  general  case,  but  they  differ  in  the 
method  of  finding  the  line  of  intersection  of  the  planes. 

In  Fig.  242  the  line  A  passes  through  the  point  s„  in 
which  HQ,  HR,   VQ   aud   VR  intersect.     A   second  point  is 


R  is  in 

quadrants 

I  and  3, 
30°  with  H 


Fig.  242. 


Fig.  243. 


obtained  by  means  of  the  auxiliary  plane  X  (Prob.  12,  Special 
Case  5,  §  118).  The  secondary  projection  Ax  passes  through 
s1  and  the  secondary  projection  of  b. 

Figure  243  is  similar  to  Figure  242  in  that  the  traces  HQ,  HR, 
VQ,  and  VR  intersect  in  a  common  point  s^  A  second  point 
b,  in  the  line  of  intersection  A,  is  here  obtained  by  means  of 
the  profile  plane  P  (Prob.  12,  Special  Case  6,  §  118),  and  the 
secondary  projection  Axv  passes  through  sx  and  the  secondary 
projection,  b^,  of  b.  In  other  respects  the  constructions  of 
Figs.  242  and  243  follow  that  of  Fig.  241. 

Special  Case  I.  The  line  of  intersection  of  the  given 
planes  is  parallel  to  H  or  V  Let  Q  and  R,  Fig.  244,  be  the 
given  planes,  with  HQ  and  HR  parallel.  Then  the  line  of  in- 
tersection, A,  is  parallel  to  H  (Prob.  12,  Special  Case  1, 
§  118).     Pass  the  auxiliary  plane  Z  perpendicular  to  A.     Since 


XVI,  §  146]    PROBLEMS  ON  THE  LINE  AND  PLANE    161 

A  is  parallel  to  H,  the  plane  Z  is  perpendicular  to  H.  The 
line  A  intersects  the  plane  Z  in  the  point  c  (cA,  c")  (Prob.  13, 
Special  Case,  §  119).  Revive  the  plane  Z  about  HZ  into  H. 
The  revolved  position,  cr,  is  obtained  simply  by  making  the 
distance  from  HZ  to  cr  equal  to  the  distance  of  C  from  GL. 
Then  it  is  evident  that  the  lines  of  intersection  of  Z  with  the 
given  planes  Q  and  R  will  take  the  positions  crs2  and  crs3,  and 


cv 

Av      \t>^^ 

>      -. 

>      ,s> 

Fig.  244. 


Fig.  245. 


the  included  angle,  0,  between  these  lines  is  the  angle  between 
the  given  planes. 

Note.  The  student  may  discover  that  TrZ,  and  also  c,  are  superfluous 
in  this  construction. 

Special  Case  II.  One  of  the  given  planes  is  parallel  to 
II  or  V.  This  case  falls  under  the  solution  just  given  ;  a  still 
simpler  solutionis  shown  in  Fig.  245.  The  given  plane  R  is 
parallel  to  II.  Therefore  the  angle  between  planes  Qand  B  is 
equal  to  the  angle,  6,  which  Q  makes  with  //,  the  latter  angle 
being  found  by  Problem  18,  §  136. 

146.  Application  of  the  Preceding  Problem.  As  instances  in 
which  the  angle  between  two  planes  is  wanted  in  actual  con- 
struction, we  may  cite  the  case  of  two  intersecting  pitch  (slop- 
ing) roof's,  or  of  two  intersecting  masonry  walls  with  battered 
(sloping)  faces.  The  way  in  which  the  problem  usually  appears 
is  as  a  corollary  to  the  following  problem. 

M 


162 


DESCRIPTIVE   GEOMETRY         [XVI,  §  146 


Problem  28.  Given  the  horizontal  traces  of  two  planes,  and  the 
angle  each  plane  makes  with  H,  to  find  the  line  of  intersection  of  the 
two  planes. 

Note.  For  the  problem  as  stated,  four  results  are  possible,  in  general. 
In  addition  to  the  above  data,  it  will  be  assumed  that  the  slopes  of  the 
planes  are  known  in  order  to  limit  the  problem  to  one  result.  This  would 
be  the  case  in  the  practical  applications  of  the  problem. 

Analysis.  One  point  in  the  line  of  intersection  is  located  at 
the  intersection  of  the  given  traces  of  the  planes.  To  locate  a 
second  point,  draw  in  each  plane  a  line  parallel  to  H,  so  that 
these  lines  will  intersect.  The  necessary  and  sufficient  condi- 
tion for  their  intersection  is  that  these  two  lines  shall  be  at  the 
same  distance  from  H. 

Construction  (Fig.  246).  Let  HQ  and  HR  be  the  given  H- 
traces.  Let  plane  Q  make  the  angle  al  with  H,  and  R  the  angle 
a2  with  H.     In  any  convenient  position,  draw  a  ground  line, 


GyLi,  perpendicular  to  HQ,  and  draw  the  F-trace  and  edge 
view,  VXQ,  of  Q,  making  the  given  angle,  a„  with  GXLX.  Draw 
a  second  ground  line,  GtL2,  perpendicular  to  HR,  and  draw  the 
F-trace  and  edge  view,  V2R,  of  R,  making  the  given  angle,  «2> 
with  Gr2L2.      Assume  any  arbitrary  distance,  d.      Draw  the 


XVI,  §  146]   PROBLEMS  ON  THE  LINE  AND  PLANE     163 

F"-trace  and  edge  view,  Vx  W,  of  a  plane  W,  parallel  to  H  and 
at  the  distance  d  above  H,  also  the  F"-trace  and  edge  view,  V2  W, 
of  this  same  plane,  that  is,  draw  V\  IF  and  V2W  parallel  respec- 
tively to  GiLx  and  G2L2,  and  at  the  assumed  distance  d.  The 
planes  Q  and  W  intersect  in  a  line  M,  one  projection  of  which 
is  the  point,  Mf,  where  FilFand  ViQ  intersect.  Project  from 
M{"  and  obtain  Mh,  parallel  to  HQ.  The  planes  R  and  W  in- 
tersect in  a  line  N,  a  projection  of  which  is  the  point  N2V,  where 
V2  IF  intersects  V2R.  Project  from  N2V  and  obtain  JVh,  parallel 
to  HR.  The  projections  Mk  and  Nh  intersect  in  bh,  which  is  an 
actual  intersection  in  space,  since  M  and  N  are  both  in  the 
plane  W.  This  point  is  common  to  both  Q  and  R.  Hence  the 
//-projection,  Ah,  of  the  required  intersection  of  Q  and  R,  is 
drawn  through  bh  and  the  intersection,  s,  of  HQ  and  HR. 

To  locate  the  line  A  further,  assume  a  third  ground  line, 
6r3L3,  coincident  with  Ah.  A  ^projection,  b3v,  of  point  b,  is 
located  by  making  the  distance  from  bzv  to  G3L3  equal  to  the 
distance  d,  since  this  is  the  distance  from  H  of  any  point  in  the 
lines  M  and  N.  The  projection,  A3V,  of  A  passes  through  b./ 
and  the  point  s.  The  line  A  is  now  definite,  since  we  have  its 
//"-projection,  Ah,  and  the  angle  «3  (between  A3V  and  G3L3)  which 
this  line  makes  with  H. 

Corollary.     To  find  the  angle  between  the  given  planes. 

In  finding  the  angle  between  two  planes  by  Problem  27,  the 
construction  already  made  in  Fig.  24G  is  a  part  of  the  construc- 
tion of  that  problem  ;  namely,  the  //-projection,  Ah,  of  the  line 
of  intersection  of  the  given  plane,  and  the  projection,  Af,  of 
this  line,  using  a  secondary  ground  line  coincident  with  Ah.  To 
complete  the  construction,  draw,  through  any  convenient  point 
o  on  Ah,  the  trace  HZ  perpendicular  to  A\  Find  the  perpen- 
dicular distance,  oc3",  from  o  to  A3V.  Locate  cr  on  Ah  by  making 
the  distance  cro  equal  to  oc3".  From  c,  draw  to  the  intersect  ions, 
s2  and  88,  of  HZ  with  HQ  and  HR  respectively.  The  angle,  0, 
between  these  lines  is  the  required  angle  between  the  planes  Q 
and  R. 


CHAPTER   XVII 
COUNTER-REVOLUTION   OF   PLANES 

147.  Counter-revolution.  A  point,  line,  or  any  number  of 
lines  lying  in  a  plane  may  be  revolved  about  one  of  the  traces 
of  the  plane  into  H  or  V.  The  basis  for  all  such  revolutions 
is  the  revolution  of  a  single  point,  given  in  Problem  21,  §  138. 

It  has  further  been  shown  (see  Prob.  22,  Corollary,  §  140 ; 
Prob.  24,  Corollary  2,  §  142  ;  Prob.  25,  §  143)  that  the  revolved 
position  of  a  line  lying  in  a  plane  can  be  given  or  assumed, 
and  the  plane  counter-revolved,  that  is,  revolved  back  about  its 
trace  so  that  the  projections  of  the  line  can  be  found. 

In  all  these  solutions,  however,  the  counter-revolution  was 
made  to  depend  upon  a  previous  direct  revolution  of  lines 
already  existing  in  the  plane.  We  shall  now  take  up  the  prob- 
lem of  a  direct  counter-revolution,  starting  only  with  the 
revolved  position  of  a  point  and  the  traces  of  the  plane ;  in 
other  words,  the  direct  reverse  of  Problem  21,  §  138. 

Problem  29.  Given  the  position  of  a  point  lying  in  a  plane  after 
the  plane  has  been  revolved  into  H  or  V  about  the  corresponding  trace, 
to  find  the  projections  of  the  point. 

Analysis.  Suppose  the  point  a,  lying  in  the  plane  Q,  to  have 
been  revolved  about  HQ  into  H.  The  figure  is  left  to  the  stu- 
dent. The  path  of  the  revolving  point  is  the  arc  of  a  circle, 
which  will  project  on  Fas  a  straight  line  perpendicular  to 
HQ,  and  on  any  plane  perpendicular  to  HQ  in  its  true  shape 
and  size.  Referring  to  the  various  solutions  of  Problem  21, 
§  138,  it  is  seen  that  the  solution  given  in  Fig.  220  shows  these 
two  projections  of  the  path  of  the  revolving  point.  The  con- 
struction of  Fig.  220  may  therefore  be  reversed  to  give  the 
solution  of  the  present  problem. 

Construction  (Fig.  247).  The  plane  Q  is  given  by  its  traces 
HQ  and  VQ,  and  ar  is  given  as  the  position  of  a  point  which 

164 


XVII,  §  147]     COUNTER-REVOLUTION  OF  PLANES     165 

has  been  revolved  about  HQ  into  H.  Assume  a  secondary 
ground  line  GXLX  perpendicular  to  HQ.  Assume  a  point,  f, 
in  VQ ;  project  to  th  in  GL.  Find  the  secondary  projection, 
txv,  of  t,  and  through  this  point  draw  the  trace  and  edge  view 
VXQ  (§  70).     Project  ar  to  GXLX.    With  o  as  center,  revolve  this 


Fig.  247. 

point  to  axv  in  VXQ.  Then  axv  is  the  secondary  projection  of 
the  point  a.  Project  from  axv  perpendicular  to  GXLX,  and  from 
ar  perpendicular  to  HQ.  Both  of  these  lines  must  pass  through 
ah,  which  is  thus  determined.  To  find  a",  we  now  have  given  a* 
as  one  projection  of  a  point  lying  in  the  plane  Q.  Through  ah 
draw  Mh  parallel  to  HQ;  this  is  one  projection  of  a  horizontal 
principal  line  of  Q.  The  other  projection,  M",  is  parallel  to 
GL;  o"  lies  in  M'  (Prob.  16,  §  133). 

Check.  The  distance  of  av  from  GL  equals  the  distance  of 
axv  from  GxIj^  (§  67). 

A  Second  Result.  The  point  a,  thus  found,  lies  above  II. 
The  given  revolved  position  may  also  he  the  revolved  position 
of  a  point  b,  lying  below  II.  To  obtain  this  result,  utter  pro- 
jecting from  the  given  revolved  position  to  GXLU  revolve  about 
o  to  bxv,  which  lies  in  V{Q  produced  below  GXLX.  Then  pro- 
ceed as  for  the  point  a.  As  before,  there  is  a  cheek  on  the 
construction  ;  the  distance  from  b"  to  GL  equals  the  distance 
from  bx"  to  GXLX. 


166  DESCRIPTIVE  GEOMETRY       [XVII,  §  147 

Note.  The  student  does  not  always  see  readily  why  the  F-projections 
av  and  bv  should  be  located  by  means  of  auxiliary  lines  in  the  plane  Q, 
since  the  distances  of  these  points  from  GL  appear  at  once  in  the  second- 
ary projection.  Indeed,  the  projections  av  and  bv  can  be  located  by  trans- 
ferring from  the  secondary  projection  the  distances  of  a\v  and  biv  from 
G\L\.  But  if  this  method  be  employed,  not  only  the  distances,  but  the 
directions  of  these  points  from  G\L\  must  be  considered,  and  there  is  a 
chance  for  error.  The  method  of  locating  a"  and  bv  by  means  of  the 
auxiliary  lines  M  and  N  is  free  from  this  ambiguity. 

In  Fig.  248,  the  given  revolved  position,  ar,  is  that  of  a  point 
which   has   been   revolved  about   VQ  into    V     A   secondary 


Fig.  248. 

ground  line,  G'L',  is  taken  perpendicular  to  VQ,  and  the  con- 
struction is  entirely  analogous  to  that  of  Fig.  247.  Two  results 
are  possible,  namely,  points  a  (av,  ah)  and  b  (bv,  &*). 

Corollary.     To  counter-revolve  a  plane  polygon. 

Analysis.  Find  the  projections  of  each  corner  of  the  polygon 
according  to  the  preceding  method. 

Construction  (Fig.  249).  The  given  revolved  position,  1,2,3  r4r, 
is  that  of  a  square  which  has  been  revolved  about  HQ  into  H. 
As  in  the  case  of  a  single  point,  this  may  be  the  revolved  posi- 
tion of  two  figures  lying  in  Q.  To  limit  the  construction  to 
one  result,  therefore,  it  is  given  that  point  1  is  the  highest 
corner  of  the  square.  The  projections  of  each  corner  of  the 
square  are  obtained  by  the  method  of  Fig.  247,  using  a  second- 
ary ground  line  perpendicular  to  HQ.     Xote  especially  point  3. 


XVII,  §  147]     COUNTER-REVOLUTION  OF  PLANES     167 

Since  the  edges  of  the  square  are  lines  lying  in  the  plane  Q, 
the  traces  of  these  lines  must  lie  in  the  traces  of  the  plane 
(§  96).  This  fact  furnishes  a  series  of  checks  on  the  work,  as, 
for  example,  that  2r3r  and  2A3h  intersect  HQ  in  the  same  point, 


namely,  the  iT-traee  of  this  line,  and  that  this  trace  projects  to 
GL  in  the  point  where  2V2>"  intersects  QL. 

Similarly,  lrIr  and  1*4*,  if  produced,  must  intersec!  HQ  in 
the  same  point.  In  fact,  the  counter-revolution  may  be  effected 
largely  by  the  use  of  the  traces  of  the  various  lines,  l>ut  ;it  leasl 
one  point  must  be  counter-revolved  as  in  the  general  problem. 

The  counter-revolution  of  a  plane  figure  is  the  reverse  of 
Problem  24,  Corollary  1,  §  142,  and  should  be  compared  with  it. 


168 


DESCRIPTIVE   GEOMETRY       [XVII,  §  148 


148.  An  Auxiliary  Problem.  As  an  application  of  the  pre- 
ceding corollary,  we  shall  construct  the  projections  of  a  right 
prism  or  a  pyramid  resting  on  an  inclined  plane,  using  the 
polygon  which  is  located  in  the  plane  as  the  base  of  the  solid  ; 
or,  what  will  result  in  the  same  construction,  we  shall  construct 
the  projections  of  a  right  prism  or  pyramid,  the  axis  of  which 
is  inclined  to  both  H  and  V.  As,  however,  the  axis  of  the 
pyramid,  or  the  long  edges  of  the  prism,  will  be  a  line  or  lines 
perpendicular  to  the  plane  of  the  base,  it  will  be  necessary  first 
to  solve  the  following  problem. 

Problem  30.  At  a  given  point  in  a  plane,  to  draw  a  line  which 
shall  be  perpendicular  to  the  plane  and  of  given  length. 

First  Analysis.  A  line  perpendicular  to  a  plane  and  indefi- 
nite in  length  may  be  drawn  by  making  the  projections  of  the 
line  perpendicular  to  the  traces  of  the  plane  (§  112).  Through 
the  given  point  draw  a  line  perpendicular  to  the  plane.  Assume 
a  second  point  on  this  line.  Find  the  true  length  of  the  line 
between  the  assumed  and  given  points.  On  this  true  length, 
measure  from  the  given  point  the  length  of  line  desired.  Ob- 
tain the  projections  of  this  line  by  reversing  the  construction 
for  finding  the  true  length. 


Construction  by  First  Analysis  (Fig.  250).  Let  a,  lying  in  Q, 
be  the  given  point.  Draw  the  line  ac  (ahch,  a1^"),  indefinite  in 
length  and  perpendicular  to  Q.     Assume  a  point  c  (c*,  C)  on 


XVII,  §  148]     COUNTER-REVOLUTION  OF  PLANES     169 

this  line,  and  find  the  true  length,  avcrv,  of  etc  (Prob.  3,  §  78). 
Measure  off  avbrv  equal  to  the  true  length  of  the  required  line. 
From  brv  find  the  projections  bv  and  bk  by  reversing  the  con- 
struction for  finding  the  true  length.  Then  ahbh  and  avb"  are 
the  projections  of  a  line  of  the  given  length. 

Note.  Any  method  of  finding  the  true  length  of  ac  may  be  reversed, 
all  leading  to  the  same  point  b. 

Second  Analysis.  Any  plane  perpendicular  to  the  given  plane 
will  be  parallel  to  the  required  line.  Hence  assume  a  secondary 
plane  of  projection  perpendicular  to  either  trace  of  the  given 
plane.  Find  the  secondary  projection  of  the  given  plane  and 
point  on  it.  Since  the  required  line  is  parallel  to  this  plane  of 
projection,  the  secondary  projection  of  the  line  may  be  drawn 
at  once,  perpendicular  to  the  secondary  trace  of  the  plane  and 
of  the  given  length.  From  the  secondary  projection  of  the  line 
the  projections  on  if  and  F~may  be  found. 

Construction  by  Second  Analysis  (Fig.  251).  Let  a,  lying  in 
Q,  be  the  given  point.     Assume  Gri-Li  perpendicular  to  HQ. 


Vui.  -T.l. 


Project  a  to  axv,  and  through  this  point  draw  the  Fi-trace  and 
edge  view,  V\Q.  Draw  <W  perpendicular  to  \r,(tK  and  make 
it  equal  to  the  given  true  length.  Draw  a*6*  perpendicular  to 
HQ;  locate  bh  by  projection  from  />,".  Project  from  l>h  to  6", 
making  the  distance  from  bv  to  GL  equal  to  the  distance  from 
b{v  to  G\LX.  Draw  avb".  As  a  check,  this  should  be  perpen- 
dicular to  VQ.    Then  a*6*  and  avbv  are  the  required  projections. 


170 


DESCRIPTIVE   GEOMETRY       [XVII,  §  149 


149.  A  Prism  or  Pyramid  Whose  Axis  Is  Inclined  to  Both  H 
and  V.  The  method  of  obtaining  the  projections  may  best  be 
shown  by  considering  concrete  examples. 

Example  1.     Draw  the  projections  of  a  right  pyramid. 

Tlie  axis  is  1|"  long,  makes  45°  with  H  and  30°  with  V,  and 
slopes  (from  the  apex  of  the  pyramid)  downward,  backward,  to 
the  right. 

The  base  is  a  regular  pentagon,  inscribed  in  a  circle  of  1" 
radius. 

Hie  lowest  edge  of  the  base  is  parallel  to  H.  The  center  of 
the  base  is  f"  above  H  and  1"  in  front  of  V. 


Scale    HaJf    Size 


Fig.  252. 


Construction  (Fig.  252).  The  first  step  in  the  construction 
is  to  find  the  plane  which  contains  the  base  of  the  pyramid. 
The  data  for  this  plane  are  not  given  directly  in  the  statement 
of  the  problem.  But,  since  the  plane  of  the  base  of  the 
pyramid  is  perpendicular  to  its  axis,  we  derive  at  once,  from 


XVII,  §  149]     COUNTER-REVOLUTION  OF  PLANES     171 

the  data  given  for  the  axis,  that  this  plane  makes  the  comple- 
mentary angles,  45°  with  H  and  60°  with  V,  and  slopes  down- 
ward, forward,  to  the  left.  Construct  the  plane  X,  using  these 
angles  and  slope  (Prob.  20,  §  137). 

We  might  next  find  a  point,  a,  in  this  plane,  distant  |" 
above  H  and  1"  in  front  of  V.  To  do  this,  draw  the  line  L, 
every  point  of  which  is  so  located  ;  that  is,  draw  If  par- 
allel to  GL  and  |"  above  it,  and  Lh  parallel  to  and  1"  in 
front  of  GL.  Find  the  point,  a,  where  L  pierces  X  (Prob.  13, 
§  119). 

In  the  figure,  however,  in  order  to  render  more  clear  the 
subsequent  steps  of  the  construction,  point  c  (c*,  c")  is  located 
at  the  given  distances  from  H  and  V,  and  through  this  point 
the  plane  Q  is  passed  parallel  to  X  (Prob.  9,  §  107). 

Revolve  c  about  HQ  into  H,  obtaining  c,  (Prob.  21,  §  138). 
"With  cr  as  center,  draw  a  circle  of  1"  radius.  Inscribe  in  this 
circle  the  regular  pentagon  lr2r3r4ror,  so  located  that  the  side 
nearest  HQ  is  parallel  to  it.  This  pentagon  is  the  revolved 
position  of  the  base  of  the  pyramid. 

Assume  a  secondary  ground  line,  GiLu  perpendicular  to  HQ. 
Project  c  to  cxv,  and  draw  VXQ  through  cf.  Using  the  edge 
view  ViQ,  counter-revolve  the  pentagon  into  the  plane  Q 
(Prob.  29,  Corollary,  §  147). 

To  obtain  the  apex  of  the  pyramid,  we  must  draw  from  c 
a  line  1  J"  long  and  perpendicular  to  Q.  Draw  from  c?  the 
line  'Y'o,",  perpendicular  to  ViQ;  make  the  length  of  this 
line  1;". 

Draw  from  c*  perpendicular  to  HQ]  project  from  of  t"  o* 
on  this  line. 

Project  from  o*  to  ov,  making  the  distances  of  ov  and  of  from 
their  respective  ground  lines  equal.  Note  as  a  check  whether 
ovcv  is  perpendicular  to  VQ.     (Compare  Fig.  251.) 

Complete  the  projections  of  the  pyramid,  making  visible 
edges  full,  and  invisible  ones  dotted. 

In  the  figure,  the  projection  of  the  entire  pyramid  on  the 
secondary  plane  of  projection  is  completed,  as  well  as  the  pro- 
jections on  //  and  V. 


172  DESCRIPTIVE   GEOMETRY       [XVII,  §  149 

Example  2.  Draw  the  projections  of  a  right  prism,  resting 
with  its  lower  base  in  a  given  plane  Q.  The  prism  is  If"  long. 
Tlie  base  is  a  square  of  1"  sides.  TJie  lowest  corner  of  the  base 
is  \"  above  H,  f "  in  front  of  V.  Two  edges  of  the  base  make 
equal  angles  with  H  and  V,  and  slope  dowmvard,  backward,  to 
the  right. 

Note.  A  plane,  like  W,  Fig.  253,  which  has  coincident  traces  parallel 
to  GL,  will  slope  downward  and  backward,  and  will  evidently  make  equal 
angles,  namely,  45°,  with  H  and  V.  Any  line  lying  in  such  a  plane  will 
make  equal  angles  with  H  and  V. 

Construction.  Let  Q,  Fig.  254,  be  the  given  plane  on  which 
the  base  of  the  prism  is  to  rest.     Draw  the  line  L  (Lv,  Lh),  \" 


Fig.  253. 

above  H  and  J"  in  front  of  V.  Find  the  point  in  which  the 
line  L  intersects  the  plane  Q  (Prob.  13,  §  119).  This  locates 
the  point  1,  the  lowest  corner  of  the  base.  Pass  through  the 
line  L  a  plane  sloping  downward  and  backward,  and  making 
45°  with  //and  V;  since  the  line  L  is  J"  above  77,  the  trace, 
HW,  is  drawn  \"  behind  Lh.  Find  the  line  of  intersection  of 
the  planes  Q  and  W.  This  is  si,  and  is  determined  by  the  point 
s,  where  HW  intersects  1IQ,  and  the  point  1  already  found. 

The  line  si  lies  in  Q,  makes  equal  angles  with  H  and  V,  and 
slopes  downward,  backward,  to  the  right.  Hence  si  determines 
one  side  of  the  base  of  the  prism.  Revolve  si  about  IIQ  into 
H  to  the  position  sHr  (Prob.  21,  Corollary,  §  138).  With  shlr 
(produced)  as  one  side,  draw  the  square  lr2r3r4r,  making  each 
side  1"  long  ;  this  square  is  the  revolved  position  of  the  base 
of  the  prism.  Note  that,  since  point  1  is  to  be  the  lowest  corner, 
no  position  of  the  square,  other  than  that  shown,  is  possible. 


XVII,  §  149]     COUNTER-REVOLUTION  OF  PLANES     173 

Assume  CrjL,  perpendicular  to  HQ;  locate  VXQ  by  means  of 
some  point,  as  e,  and  counter-revolve  the  square  (Prob.  29, 
Corollary,  §  147).  As  a  check,  note  that  the  point  2  must  fall 
on  the  line  si,  produced,  already  determined. 

The  long  edges  of  the  prism  are  lines  If"  long  and  perpen- 
dicular to  Q.  Draw  these  lines,  lxv  5^,  21r61",  etc.,  in  the 
secondary  projection  ;  from  this  projection  find  the  projections 
on  H  and  V  (Prob.  30,  Second  Analysis,  §  148).  Complete  the 
projections  of  the  prism  as  shown. 


An  alternative  method  for  finding  the  long  edges  of  the 
prism  is  also  given  in  Fig.  254.  From  point  1  draw  the  line 
(lhfh,  L'/'j  perpendicular  to  Qand  of  unknown  length.  Find 
the  true  length  1"./V  Make  1*5! the  required  length,  L§";  from 
5t  find  the  projections  5' and  5*  (Prob.  30,  First  Analysis,  >  1  18). 
The  remaining  long  edges,  2-6,  3-7,  and  4-8,  may  now  be 
drawn  equal  and  parallel  to  1-5. 


CHAPTER   XVIII 


TANGENT    LINES    AND    PLANES  —  GENERAL    PRINCIPLES 


150.  Curves.  A  curve  may  be  defined  as  a  line,  no  portion  of 
which  is  straight. 

A  plane  curve  is  one  which  lies  wholly  in  a  plane.  If  no 
part  of  the  curve  is  plane,  the  curve  is  a  space  curve,  or  curve  of 
three  dimensions.  Familiar  examples  of  plane  curves  are  circles 
and  ellipses.  An  example  of  a  space  curve  is  the  helix,  as 
shown  by  a  screw  thread  or  a  spiral  spring.  Space  curves 
result  usually,  though  not  necessarily,  when  two  curved  sur- 
faces of  any  kind  intersect  each  other. 

151.  Projections  of  Curves.  The  projection  of  a  plane  curve 
on  any  plane  of  projection  is,  in  general,  a  curve.  If,  however, 
the  plane  of  the  curve  is  perpendicular  to  the  plane  of  projec- 
tion, the  projection  of  the  curve  is  a  straight  line.  The  pro- 
jection of  a  space  curve  on  any  plane  is  always  a  curve. 

Conversely,  if  one  projection  of  a  curve  is  a  straight  line, 
the  curve  is  a  plane  curve.  But  if  both  projections  of  a  curve 
are  curves,  the  actual  curve  may  or  may  not  be  a  plane  curve, 

so  that  no  general  rule  can  be  given. 

152.  Tangent  Lines.  At  any  point 
in  a  curve,  a  straight  line  tangent  to 
the  curve  can  be  drawn.  The  pro- 
jection on  any  plane  of  the  curve 
and  its  tangent  are  tangent  to  each 
other.  (See  Fig.  255.)  An  excep- 
tion occurs  if  the  plane  of  projection 
is  perpendicular  to  the  rectilinear 
tangent,  since  the  latter  then  pro- 
jects as  a  point. 
174 


Fig.  255. 


XVIII,  §  154]      TAXGEXT  LIXES  AXD  PLANES 


175 


153.  Tangents  to  Plane  Curves.  A  straight  line  tangent  to  a 
plane  curve  lies  in  the  plane  of  the  curve. 

Hence,  if  it  is  required  to  draw  a  straight  line  tangent  to  a 
plane  curve,  from  a  point  not  on  the  curve,  there  is  no  solution 
unless  the  given  point  lies  in  the  plane  of  the  curve.  In  other 
words,  a  straight  line  tangent  to  a  plane  curve  cannot  be  drawn 
from  a  point  in  space  which  lies  outside  the  plane  of  the  curve. 


Fig.  256. 


Fig.  257. 


For  example,  Fig.  256  represents  the  projections  of  a  plane 
curve  C  (§  151),  and  of  two  points  a  and  b.  A  line  tangent 
to  the  curve  C  cannot  be  drawn  from  the  point  a,  since  it  is 
obvious  that  a  is  not  in  the  plane  of  the  curve.  Why  ?  On 
the  other  hand,  the  point  b  lies  in  the  plane  of  the  curve,  and 
from  it  two  tangents  (not  shown  in  the  figure)  may  be  drawn 
to  the  curve  C. 

154.  Tangent  Planes  to  Curved  Surfaces.  Let  S,  Fig.  257, 
represent  any  curved  surface.  Let  a  be  any  point  on  the  sur- 
face, and  let  T  be  the  tangent  plane  to  the  surface  at  the 
point  a. 

Let  Cbe  any  curve  drawn  on  the  surface  through  the  point 
a.  If  drawn  at  random,  C  will  probably  be  a  space  curve,  al- 
though it  may  be  taken  so  as  to  be  a  plane  curve.  Let ./  be 
the  line  tangent  to  the  curve  C  at  the  point  a.  Then  ./  will 
be  tangent  to  the  surface  8,  and  will  lie  in  the  tangenl  plane 
T.    Let  D  be  a  second  curve  lying  on  &  and  passing  through 


176  DESCRIPTIVE  GEOMETRY       [XVIII,  §  154 

a.     Then   a   line  K,  tangent   to  D  at  a,  will  also  lie  in  the 
plane  T. 

It  follows  that  the  tangent  plane  to  a  surface  at  a  point  on 
it  may  be  defined  as  the  plane  which  contains  all  the  straight 
lines  tangent  to  the  given  surface  at  the  given  point  of  the 
surface. 

155.  Determination  of  Tangent  Planes  by  Means  of  Tangent 
Lines.  Two  intersecting  straight  lines  determine  a  plane. 
Suppose  the  given  surface  to  be  such  that  two  curves  of  known 
properties  can  be  drawn  on  the  surface  through  the  required 
point  of  tangency.  The  tangent  plane  is  then  determined  by 
finding  the  plane  which  contains  the  two  rectilinear  tangents  to 
these  curves.    In  general,  the  curves  used  will  be  plane  curves. 

If  the  surface  is  such  that  a  straight  line,  as  E,  Fig.  257,  can 
be  drawn  on  the  surface  through  the  given  point  a,  the  line  E 
will  lie  in  the  tangent  plane  T.  Such  a  straight  line  which 
lies  wholly  on  the  surface  is  called  a  rectilinear  element  of  the 
surface.  In  this  case,  but  one  tangent  line  in  addition  to  the 
element  E  is  necessary  to  determine  the  plane  T. 

Curved  surfaces  exist  in  which  two  rectilinear  elements  of 
the  surface  can  be  drawn  through  a  given  point  in  the  surface. 
For  such  surfaces  the  tangent  plane  at  any  point  is  the  plane 
containing  the  two  rectilinear  elements  passing  through  the 
point. 

156.  The  Normal.     The  normal  to  a  surface  at  any  point  on 

it  is  the  straight  line  perpendicular 
to  the  tangent  plane  at  that  point. 
For  certain  curved  surfaces,  for 
example,  the  sphere  shown  in  Fig. 
258,  the  normal  can  be  determined 
readily  from  known  properties  of 
the  surface.  Hence  the  tangent 
plane  T  at  the  point  a  can  be  found 
by   drawing   first  the   normal  N, 

and    then    passing    through    the 
point  a  the  plane  T  perpendicular  to  the  line  N. 


XVIII,  §  157]     TANGENT  LINES  AND  PLANES 


177 


157.  Determination  of  Tangent  Planes  by  Means  of  the  Normal. 
It  is  to  be  noted  that  the  tangent  plane  T  is  absolutely  deter- 
mined in  space  by  means  of  the  normal  N,  since  but  one  plane 
can  be  passed  through  a  given  point  a  perpendicular  to  a  given 
line  N.  But  in  a  projection  drawing,  where  the  plane  T  is 
not  determined  until  its  traces  HT  and  VT  are  found,  addi- 
tional lines  are  necessary.  This  is  shown  in  Fig.  259.  The 
point  a  represents  the  given  point  in 
some  given  surface,  and  the  line  N  rep- 
resents the  normal  at  that  point.  The 
tangent  plane  T  is  passed  through  the 
point  a  perpendicular  to  N.  But  al- 
though the  projections  Nh  and  Nv  give 
the  directions  of  HT  and  VT  respec- 
tively (§  112),  the  normal  alone  fur- 
nishes no  point  on  either  of  these  traces. 
Hence  at  least  one  line,  as  J,  lying  in 
the  tangent  plane,  is  required  in  the  projection  drawing  to 
supply  a  necessary  point,  as  s.     (See  Prob.  10,  §  115.) 

In  determining  the  traces  of  a  tangent  plane  by  the  use  of 
the  normal,  therefore,  the  working  method  consists  in  finding, 
in  addition  to  the  normal,  one  line  which  is  tangent  to  the 
given  surface.  The  required  tangent  plane  is  then  passed 
through  the  tangent  line  perpendicular  to  the  normal. 


Fig.  259. 


CHAPTER   XIX 

TANGENT  PLANES  TO  CONES  AND  CYLINDERS 

158.  The  Cone  and  Cylinder.  Definitions.  The  terms  cone  and 
cylinder  are  variously  used,  both  in  mathematical  and  in  popu- 
lar language,  to  denote  either  surfaces,  or  solids  bounded  in 
part  by  these  surfaces. 

As  a  surface,  a  cone  may  be  defined  as  generated  by  a  mov- 
ing straight  line,  indefinite  in  extent,  which  always  passes 
through  a  fixed  point  in  space,  and  is  so  guided,  by  a  curved 
line  or  otherwise,  that  a  plane  is  not  formed.  A  cylinder  dif- 
fers from  a  cone  in  that  the  generating  straight  line,  instead 
of  passing  through  a  fixed  point,  remains  always  parallel  to 
its  initial  position.  In  both  of  these  surfaces,  any  position  of 
the  generating  straight  line  is  known  as  a  rectilinear  element, 
or  simply  as  an  element,  of  the  surface. 

The  solids  understood  by  the  terms  cone  and  cylinder  are 
familiar  objects,  and  do  not  need  to  be  defined  here.  It 
should  be  remembered,  however,  that  the  axis  of  a  cone  or 
cylinder  is  not  necessarily  at  right  angles  to  the  base. 

159.  Representation  of  the  Cone  and  the  Cylinder.  We  shall 
consider  at  present  only  the  solid  forms  of  these  objects,  and 
shall  confine  ourselves  to  those  in  which  the  base  is  either  a 
circle  or  an  ellipse.  The  projections  (Ch,  C")  of  a  general 
cone,  in  which  the  base  is  an  ellipse,  and  the  axis  oblicpie  to 
the  plane  of  the  base,  are  given  in  Fig.  2C0. 

We  shall  not  attempt  to  solve  problems,  however,  when  the 
cone  is  projected  as  in  Fig.  260.  Since  the  base  of  the  cone 
is  a  plane  curve,  it  can  be  projected  as  a  straight  line  (§  151). 
In  Fig.  260,  let  B  (HB,  VB)  be  the  plane  of  the  base,  found  by 
passing  a  plane  through  any  three  points  of  the  curve  (Prob. 
6,  Cor.  II,  §  106 ;  construction  not  shown).     Assume  a  second- 

178 


XIX,  §  160] 


TANGENT  PLANES 


179 


ary  ground  line  perpendicular  to  HB ;  then  the  plane  B  will 
project  edgewise  as  ViB  (see  Fig.  92,  §  70).     Consequently,  in 


Fig.  200. 


the  secondary  projection,  CV,  of  the  cone,  the  base  will  pro- 
ject as  a  straight  line. 

Since  this  transformation  always  can  be  effected,  we  shall 
discuss  further  only  those  cases  of  cones  and  cj'linders  in 
which  one  projection,  at  least,  of  the  base  is  a  straight  line. 
This  results  when  the  plane  of  the  base  is  perpendicular  to, 
or  coincident  with,  one  of  the  coordinate  planes. 

160.  Projections  of  Cones  and  Cylinders.  We  shall  place 
these  objects  in  the  first  or  third  quadrants  only  (see  §  19). 
Projections  of  cones  are  given  in  Figs.  201-265.  These  fig- 
ures represent  the  following  objects  : 

Fig.  2G1.  Cone  in  the  third  quadrant;  base  in  a  plane  per- 
pendicular  to  V;  base  visible.  Note  that  if  we  attempt  to  read 
these  as  the  projections  of  a  cone  in  the  first  quadrant,  the 
base  should  be  invisible.  As  this  is  not  the  case,  the  coin' 
cannot  be  in  the  first  quadrant.  In  this  and  similar  cases, 
the  student  will  often  be  expected  to  determine  the  quadrant 
from  the  given  visibility  of  the  projection. 


180 


DESCRIPTIVE  GEOMETRY         [XIX,  §  160 


Fig.  262.     Cone  in  the  third  quadrant;  base  in  a  plane  per- 
pendicular to  H  ;  base  invisible. 


Fig.  261. 


Fig.  262. 


Fig.  263.  Cone  in  the  first  quadrant;  base  in  H;  base 
invisible. 

Fig.  264.  Cone  in  the  third  quadrant;  base  in  V;  base 
visible. 


Fig.  263. 


Fig.  264. 


Fig.  265.     Cone  of  revolution;  base  a  circle  in  the  P-plane. 
While   these   are   the   projections   resulting   from   projecting 


XIX,  §  160] 


TANGENT  PLANES 


181 


this  object,  so  placed,  they  are,  by  themselves,  and  without 
further  information,  ambiguous.  They  should  be  supple- 
mented by  the  profile  projection,  and 
the  quadrant  in  which  the  object  lies 
must  be  indicated  in  some  manner,  as 
for  example,  by  lettering  the  vertex. 

In  projecting  a  cylinder,  only  one 
base  will,  in  general,  be  shown,  the 
cylinder  being  left  indefinite  in  ex- 
tent in  one  direction.  Cylinders  are 
projected  in  Figs.  266-270,  which  rep- 
resent the  following  objects  : 

Fig.  266.     Cylinder  in  the  third  quad- 
rant; base  in  a  plane  perpendicular  to 
V;  base  visible.     As  in  the  correspond- 
ing case  of  the   cone    (Fig.   261),   an  Fl«-  265. 
attempt  to  read  these  views  as  the  projections  of  a  cylinder 
in  the  first  quadrant  fails. 


Fig.  266. 


Fig.  267. 


Fig.  267.     Cylinder  in  the  third  quadrant ;    base   in  a  plane 
perpendicular  to  II;  base  invisible. 


182 


DESCRIPTIVE  GEOMETRY         [XIX,  §  160 


Eig.  268.  Cylinder  in  the  first  quadrant ;  base  in  V;  base 
invisible. 

Fig.  269.  Cylinder  in  the  third  quadrant ;  base  in  H ;  base 
visible.     The  elements  of  this  cylinder  are  all  parallel  to  V. 


Fig.  268. 


Fig.  269. 


Fig.  270. 


Fig.  270.  Cylinder  of  revolution ;  base  a  circle  lying  in  P. 
Like  the  cone  of  revolution  (Fig.  265),  these  projections  are 
ambiguous,  and  similar  re- 
marks apply. 

161.  Projection  of  a  Point  in 
the  Surface  of  a  Cone  or  Cylin- 
der. Only  one  projection  of 
a  point  lying  in  a  conical  or 
cylindrical  surface  can  be  as- 
sumed. The  other  projection 
of  the  point  may  be  found 
by  means  of  an  element  of 
the  surface,  as  follows. 

Let  a  cone  be  given  as  in 
Fig.  271,  and  let  av  be  an  as- 
sumed   projection.     The   ele- 
ment E,  passing  through  a,  Fro-  271. 
must  also  pass  through  the  vertex  o.     Hence  draw  E"  through 
av  and  o".     Produce  Ev  to  meet  the  F-projection  of  the  base 


XIX,  §  161] 


TAXGEXT  PLAXES 


18.; 


at  c".  Project  from  c"  to  c*  in 
the  if-projection  of  the  base. 
Then  Eh  is  drawn  by  connect- 
ing c*  and  o*.  Find  a*  in  Eh 
by  projecting  from  a".  A 
second  result  is  possible,  namely, 
point  b  lying  in  the  element  F. 

In  general,  an  assumed  pro- 
jection will  represent  two  points. 
But  let  kv  be  assumed  in  the 
outside  or  contour  element  L. 
Then  kv  is  the  projection  of 
but  one  point  in  the  surface. 

A  similar  construction  for  the 
cylinder  is  shown  in  Fig.  272, 
the  assumed  projections  being 
the  double  point  arbv,  and  the 
single  point  kv. 

In  Fig.  273,  the  assumed  projections  are  ah,  bh.  The  base 
of  the  cone  lies  in  P,  which  necessitates  the  use  of  the  P-pro- 
jection  of  the  base.  (Compare  Fig.  265.)  A  similar  construc- 
tion applies  to  a  cylinder  whose  base  lies  in  P. 


Fig.  272. 


184  DESCRIPTIVE  GEOMETRY         [XIX,  §  162 

162.  Tangent  Planes  to  Cones  and  Cylinders.  At  every  point 
in  the  surface  of  a  cone  or  cylinder,  a  plane  tangent  to  the  sur- 
face can  be  drawn. 

Planes  tangent  to  cones  and  cylinders  can  also  be  passed  to 
fulfill  certain  other  conditions.  Thus,  the  tangent  planes  may 
be  required  to  contain  a  given  point  outside  the  surface,  or 
to  be  parallel  to  a  given  straight  line. 

In  determining  these  tangent  planes,  we  shall  make  use  of 
the  following  propositions. 

(a)  Through  every  point  in  the  surface  of  a  cone  or  cylinder, 
a  rectilinear  element  can  be  drawn,  and  this  element  will  lie 
in  the  tangent  plane  at  that  point  (jj  155). 

It  is  a  property  of  both  the  cone  and  cylinder  that 
a  plane  tangent  at  any  point  in  a  given  element  is  tangent 
at  every  point  in  the  element.  Hence,  if  the  element  of  tan- 
geney  is  known,  a  second  line  in  the  tangent  plane  may  be 
drawn  tangent  to  the  cone  or  cylinder  at  any  point  in  this 
element.  A  convenient  point  is  generally  the  point  in  which 
the  element  of  tangency  intersects  the  base. 

(c)  If  the  base  of  a  cone  or  cylinder  lies  in  H.  any  line  tan- 
gent to  the  base  lies  in  H  (§  153),  and  thus  becomes  the  H- 
trace  of  a  plane  tangent  to  the  surface.  Similarly  if  the  base 
lies  in  Vot  P. 

(d)  Every  plane  tangent  to  a  cone  contains  the  vertex. 

Problem  31.  To  pass  a  plane  tangent  to  a  cone  at  a  given  point 
in  the  surface. 

Analysis.  The  plane  is  determined  by  the  element  which 
passes  through  the  given  point,  and  a  line  tangent  to  the  base 
at  the  point  where  this  element  intersects  the  base  (§§  155, 
162,  &). 

Construction.     Case  I.      77/<=  base  of  the  cone  lien  in  H  or  V. 

Example  1  (Fig.  274).  The  base  of  this  cone  lies  in  H. 
Let  a  (oh.  a'),  lying  in  the  element  E  (Eh.  E*)  be  the  given  point 
in  the  surface.  Since  the  required  tangent  plane  contains  the 
element  E,  find  the  traces  sx  and  t-,  of  E.  Then  the  Z7-traee. 
HQ.  of  the  tangent  plane  passes  through  slf  and  is  determined 


XIX,  §  162] 


TANGENT  PLANES 


185 


by  the  fact  that  HQ  must  be  tangent  to  the  base  of  the  cone 
(§  162,  c).  The  trace  VQ  is  now  determined,  since  it  must 
pass  through  tx  and  the  point  in  which  HQ  intersects  GL. 

Let  it  also  be  required  to  find  the  tangent  plane  at  the  point 
b,  lying  in  the  element  F.     The  Zf-trace,  HT,  of  this  plane, 


Fig.  274. 

passes  through  the  //-trace,  s2,  of  F,  and  is  tangent  to  the  base 
of  the  cone.  The  F-trace,  VT,  passes  through  the  V-trace  of  F, 
but  this  point  is  inaccessible.  Recourse  must  be  had  to  some 
auxiliary  line  lying  in  the  plane  T,  and  so  situated  that  its 
F-trace  is  accessible.  (See  §  108.)  As  HT  is  known,  a  hori- 
zontal principal  line  of  the  plane  may  be  drawn  through  any 
point  of  the  line  F  (§  108,  Ex.  3).  A  convenient  point  is  the 
given  point  b.  Through  6*  draw  U  parallel  to  HT;  through 
bv  draw  Lv  parallel  to  GL.  Then  the  line  L  lies  in  the  plane 
T,  and  the  F-trace,  t2,  of  L  is  a  point  in  VT. 

'Example  2   (Fig.  275).     The  base  of  this   cone  lies  in    V. 
Tangent  planes  are  passed  at  the  points  a  and   b.     For   the 


186 


DESCRIPTIVE  GEOMETRY        [XIX,  §  162 


plane  Q,  tangent  at  a,  the  trace  VQ  is  drawn  tangent  to  the 
base  of  the  cone  through  the  F-trace  of  the  element  E.  Since 
the  i7-trace  of  the  element  E  is  not  available,  a  point  on  HQ  is 
found  by  means  of  the  auxiliary  line  L,  a  vertical  principal 


Fig.  275. 


line  of  Q,  drawn  through  the  given  point  a.     The  77-trace  of  L 
lies  in  HQ. 

The  plane  T  is  tangent  at  the  point  b.  The  trace  VT  is 
drawn  tangent  to  the  base  of  the  cone,  through  the  "F-trace  of 
the  element  F  passing  through  b.  In  this  case  VT  does  not 
intersect  GL  within  reach,  neither  is  the  iJ-trace  of  F  obtain- 
able. Hence  two  points  on  HT  must  be  obtained  by  the  use  of 
auxiliary  lines.  Two  convenient  lines  are  the  vertical  principal 
lines,  M  and  N,  of  T,  drawn,  one  through  the  given  point  b, 
the  other  through  the  vertex,  o,  of  the  cone,  since  the  vertex  is 
in  every  tangent  plane  (§  162,  d).  The  .ff-traces,  s4  and  s5,  of 
these  two  lines  determine  HT. 


XIX,  §  162] 


TANGENT  PLANES 


187 


Case  II.  Hie  base  of  the  cone  lies  in  P  (Fig.  276).  The  base 
of  this  cone  is  a  circle,  and  the  cone  is  the  symmetric  form 
known  as  the  cone  of  revolution.  Let  a  (ah,  av)  be  the  given 
point.  Point  a  lies  in  the  element  E,  which  intersects  the 
plane  of  the  base  in  the  point  c,  the  profile  trace  of  the  line. 


Through  the  actual  trace  cp  draw  the  profile  trace,  PQ,  of  the 
tangent  plane,  tangent  to  the  profile  projection  of  the  base. 
From  PQ  are  found  the  points,  t2  on  VQ,  and  s2  on  HQ  (§  60). 
.Since  E  lies  in  Q,  find  also,  if  possible,  the  II-  and  F^traces 
oi  E.  In  this  case  the  P~-trace,  tu  may  be  found,  and  this  is 
sufficient;  for  VQ  passes  through  t2  and  tu  and  1IQ  through 
s2  and  the  point  in  which  VQ  intersects  OL. 

The  plane  Tis  tangent  at  the  point  !>,  lyin^  in  the  element  F. 
This  element  intersects  <!)'■  base  at  d.  The  profile  I  race,  /''/'.  is 
drawn  through  dp  tangenl  to  the  base.  From  Prare  obtained 
the  points  *3  on  HT and  t3  on  FT.  Neither  the  //-  aoi  V  trace  of 
F  is  accessible,  so  that  an  auxiliary  Line  is  necessary.  Since 
the  vertex  of  the  cone,  o,  is  in  every  tangenl  plane  (i'   L62,  d)t 


188 


DESCRIPTIVE  GEOMETRY         [XIX,  §  162 


let  an  auxiliary  profile  plane,  X,  be  passed  through  o.  Since 
X  is  parallel  to  P,  it  will  intersect  the  plane  T  in  a  line,  L, 
parallel  to  PT.  Project  o  to  the  plane  P,  and  find  the  profile 
projection,  op.  Through  op  draw  the  P-projection,  Lp,  of  the 
line  L,  parallel  to  PT.  Find  the  traces  of  L ;  first  on  P,  then 
by  projection  to  X,  giving  the  .Ef-trace,  s^  and  the  F-trace,  t4. 

The  traces  HT 
and  VT  are  now 
determined,  since 
we  have  two 
points  on  each. 

Case  III.  TJie 
base  of  the  cone 
does  not  lie  in  H, 
V,  or  P.  In  this, 
the  general  case 
of  this  and  the 
succeeding  prob- 
lems, one  pro- 
jection of  the 
base  will  always 
be  taken  as  a 
straight  line. 
(See  §  159.)  The 
base  of  the  given 
cone,  Fig.  277,  lies  in  a  plane  perpendicular  to  H  and  oblique 
to  V.  Let  a  be  the  given  point  of  tangency.  This  point 
lies  in  the  element  E,  which  intersects  the  base  in  the  point  c. 
The  plane  tangent  at  a  is  also  tangent  at  c ;  hence  through  c 
draw  the  line  J,  tangent  to  the  base  of  the  cone  (§  162,  b). 
The  required  tangent  plane,  Q,  is  now  determined  by  the  lines 
E  and  J.  In  this  case  a  sufficient  number  of  the  traces  of  E 
and  J  may  be  found  to  determine  the  traces  HQ  and  VQ  (Prob. 
6,  §  106). 

The  plane  tangent  at  the  point  b  will  contain  the  element  F 
passing  through  b.  Line  F  intersects  the  base  at  d.  Through 
d  draw  the  line  K  tangent  to  the  base  ;  then  K  is  a  second  line 


Fig.  276  (repeated). 


XIX,  §  162]  TANGENT  PLANES  189 

in  the  tangent  plane  at  b  (§  162,  b).  But  in  attempting  to  pass 
a  plane  through  F  and  K,  only  the  traces,  s3  and  t3,  of  K 
can  be  found ;  neither  trace  of  F  is  available.  Therefore, 
choose  any  point  in  F;  a  convenient  point  is  the  vertex,  o. 


Fig.  277. 

Through  o  draw  the  auxiliary  line  L,  parallel  to  K\  then  L 
lies  in  the  plane  of  F  and  K  (§  108,  Ex.  1).  Line  L  supplies 
the  traces  s4  and  tt.  These,  taken  in  connection  with  the  traces 
of  K,  are  sufficient  to  locate  the  traces,  HT  and  VT,  of  the 
required  tangent  plane. 

Problem  32.  To  pass  a  plane  tangent  to  a  cone  through  a  given 
point  without  the  surface.     (Two  results.) 

Analysis.  Pass  a  line  through  the  given  point  and  the 
vertex  of  the  cone.  Find  the  point  in  which  this  line  pierces 
the  plane  of  the  base  of  the  cone.  From  this  piercing  point, 
draw  a  line  tangent  to  the  base  of  the  cone.  The  required 
tangent  plane  is  determined  by  the  tangent  line  and  the  line 
first  drawn.  In  general,  two  tangents  may  be  drawn  to  the 
base,  giving  two  possible  tangent  planes. 

To  prove  this  analysis,  note  that  the  vertex  of  the  cone  lies 
in  every  tangent  plane  (§  162,  d).  Hence  the  Line  firsl  drawn 
must  lie  in  the  required  tangent  plane.     A  tangent  to  the  base 


190  DESCRIPTIVE  GEOMETRY         [XIX,  §  162 

which  intersects  this  line  must  also  lie  in  the  required  tangent 
plane.  But  a  line  tangent  to  the  base  must  lie  in  the  plane  of 
the  base,  hence  it  can  be  drawn  only  from  that  point  of  the 
first  line  in  which  it  intersects  the  plane  of  the  base  (§  153). 

If  the  base  of  the  cone  lies  in  H,  the  point  in  which  the  line 
connecting  the  given  point  with  the  vertex  pierces  the  plane  of 
the  base  becomes  the  i/-trace  of  this  line,  while  the  tangent 


Fig.  278. 


drawn  from  this  trace  becomes  the  if-trace  of  the  tangent 
plane  (§  162,  c).  The  construction  is  similar  if  the  base 
lies  in  V. 

Construction.  Case  I.  TJie  base  of  the  cone  lies  in  H  or  V 
(Fig.  278).  The  base  of  the  cone  here  given  lies  in  V,  and  a 
(ah,  av)  is  the  given  point.  Draw  the  line  B  (Bh,  B0),  connect- 
ing a  and  the  vertex  o.  Find  the  traces,  s  and  t,  of  B.  Since 
the  base  of  the  cone  lies  in  V,  draw  VQ  tangent  to  the  base 
from  the  F-trace,  t,  of  B.  The  trace  HQ  passes  through  the 
#-trace,  s,  of  B  (§  98).  A  second  tangent  plane  T  (HT,  VT) 
is  similarly  obtained. 


XIX,  §  162] 


TAXGEXT  PLAXES 


191 


Cask  II.  TJie  base  of  the  cone  does  not  lie  in  H  or  V 
(Fig.  279).  The  base  of  the  cone  here  given  lies  in  the  plane 
X,  perpendicular  to  V.  The  given  point  is  a.  Draw  the  line 
B,  connecting  a  with  the  vertex  of  the  cone.  Find  the  point, 
c,  in  which  B  pierces  X.     (See  Fig.  178,  §  119.)     From  c  draw 


Fig.  279. 


the  line  J  (Jh,  Jv)  tangent  to  the  base  of  the  cone.  The 
required  tangent  plane,  Q,  may  now  be  passed  through  the 
lines  B  and  ,/  (Prob.  6,  §  10(5).  For  this  plane  no  auxiliary 
lines  are  needed.  Draw  also  from  c  the  tangenl  line  A' 
(A"A,  A"").  Then  a  second  tangent  plane  is  determined  by  the 
lines  B  and  K.  For  this  plane,  the  fl"-trace,  //'/'.  is  deter- 
mined by  the  traces,  s{  and  8i}  of  B  and  K,  respectively.  <  hie 
point  of  VT  is  £3,  the  F"-trace  of  K.  A  second  point  is  /,, 
which  is  found  by  using  the  auxiliary  line  L,  drawn  through 
the  vertex  of  the  cone  parallel  to  HT  (§  L08,  Ex.  .'}). 


192 


DESCRIPTIVE  GEOMETRY         [XIX,  §  162 


Problem  33.  To  pass  a  plane  tangent  to  a  cone  parallel  to  a 
given  line.     (Two  results.) 

Analysis.  The  required  tangent  plane  must  contain  the  ver- 
tex of  the  cone  (§  162,  d).  Moreover,  since  it  is  to  be  parallel 
to  a  given  line,  the  plane  must  contain  a  line  parallel  to  the 
given  line  (§   105).     Hence,  through  the  vertex  of   the  cone 


Fig.  280. 


pass  a  line  parallel  to  the  given  line.  Pass  the  required 
tangent  planes  through  this  line.  For  the  method  of  accom- 
plishing this,  see  the  preceding  Problem. 

Construction.  Case  I.  The  base  of  the  cone  lies  in  H  or  V 
(Fig.  280).  The  base  of  the  given  cone  lies  in  V.  The  given 
line  is  A.  Through  the  vertex,  o,  of  the  cone,  draw  B  (Bh,  B") 
parallel  to  A.  The  F-trace  of  B  is  tv  From  tx  draw  VQ  tan- 
gent to  the  base  of  the  cone.  The  trace  HQ  passes  through 
the  i?-trace,  sx,  of  B.  For  a  second  result,  from  tx  draw  VT 
tangent  to  the  base  of  the  cone.      Since  VT  does  not  intersect 


XIX,  §  162] 


TANGENT  PLANES 


193 


GL  within  reach,  an  auxiliary  line  is  needed  to  locate  HT. 
But  VT  is  so  nearly  parallel  to  GL  that  the  use  of  a  vertical 
principal  line  of  T  will  not  give  an  especially  accurate  con- 
struction. Let  us,  then,  note  the  point  of  tangency,  tz,  of  VT 
and  the  base.      Through  t2  draw  the  element  E(EV,  Eh),  the 


Fig.  281 


element  of  tangency,  which  must  lie  in  the  plane  T.  Hence 
the  //-trace,  s2,  of  E  lies  in  JIT,  which  is  now  determined  by  s2 
and  the  //-trace,  su  of  B. 

Case  II.  The  base  of  the  cone  does  vot  lie  in.  II  or  V 
(Fig.  281).  The  base  of  the  given  cone  lies  in  the  plane  X, 
perpendicular  to  V.  The  given  line  is  J.  Through  the  ver- 
tex, o, of  the  cone,  draw  />'  parallel  to  A.  Line  11  intersects 
the  plane  X  in  the  point  c.  From  c  draw  the  tangents,  J* and 
K,  to  the  base  of  the  cone.  Plane  (J  is  passed  through  B  and 
the  tangent ./,  while  plane  T  is  passed  through  />'  and  the  tan- 
gent 1\.  No  auxiliary  lines  arc  necessary.  The  planes  Q  and 
Tare  the  required  tangent  planes. 

o 


194 


DESCRIPTIVE  GEOMETRY         [XIX,  §  162 


Note.  It  will  be  seen  in  Figs.  280  and  281  that  if  the  line  B,  parallel 
to  A,  intersects  the  plane  of  the  bases  of  the  cone  within  the  circumfer- 
ence of  the  base,  no  tangent  lines  will  be  possible.  Consequently,  there 
will  be  no  solution.  This  will  happen  when  the  line  A  is  parallel,  or 
nearly  so,  to  the  axis  of  the  cone. 

A  special  case,  in  which  one,  and  only  one,  solution  exists,  will  occur 
when  the  line  B,  parallel  to  A,  is  found  to  coincide  with  an  element  of 
the  given  cone. 

Problem  34.  To  pass  a  plane  tangent  to  a  cylinder  at  a  given 
point  in  the  surface. 

Analysis.  As  in  the  corresponding  problem  with  the  cone 
(Prob.  31),  the  tangent  plane  is  determined  by  the  element 


which  passes  through  the  given  point,  and  a  line  tangent  to  the 
base  at  the  point  where  this  element  intersects  the  base. 

Construction.  Case  I.  The  base  of  the  cylinder  lies  in  H  or  V 
(Tig.  282).  The  construction  is  entirely  similar  to  that  for  the 
corresponding  case  of  the  cone  (Prob.  31,  Case  I),  and  it  should 
not  require  an  extended  explanation. 

The  plane  Q,  tangent  at  the  point  a,  contains  the  element  E. 
No  auxiliary  line  is  needed. 


XIX,  §  162] 


TANGENT  PLANES 


195 


The  plane  T  is  tangent  at  the  point  b,  lying  in  the  element 
F.  A  necessary  point,  t3,  in  VT,  is  here  determined  by  the 
auxiliary  line  L. 

Case  II.  The  base  of  the  cylinder  lies  in  P  (Fig.  283).  The 
given  cylinder  is  a  cylinder  of  revolution,  with  a  circular  base 
lying  in  P.  Obtain,  if  not  already  given,  the  profile  projection 
of  the  base.     For  the  plane  Q,  tangent  at  the  point  a,  obtain 


s 

Fh 

HQ. 

a. 

i 

uh" 

1 

\    1      »' 

1 

\ 

>^r 

' 

\ 
\ 

VQ. 

\!a* 

Ev 

t 

uvyup        >. 

VT 

/         •         \ 
-  - (       -+         J 

Fv 

bvJ 

^\j      y 

HT 

-. 

(L 
-  _       > 

\<> 

Fig.  28 

; 

the  profile  trace,  np,  of  the  element  E  containing  a.  Through 
uv  draw  the  profile  trace,  PQ,  tangent  to  the  base.  This  trace 
furnishes  the  point  s  on  HQ  and  the  point  /  on  VQ.  In  this 
case  no  other  points  are  necessary,  since  the  plane  (J  is 
evidently  parallel  to  GL.  The  plane  T  (PT,  HT,  VT),  tan- 
gent at  the  point  b  in  the  element  F,  is  obtained  in  a  similar 
manner. 

Case  III.  Tlie  base  of  the  cylinder  does  not  lie  in  H,  V,  or 
P  (Fig.  284).  The  construction  is  entirely  similar  to  thai  Eor 
the  corresponding  case  of  the  cone  (Prob.  31,  Case  III),  and 
should  be  compared  with  it.  A  detailed  explanation  will  Dot 
be  given. 


196 


DESCRIPTIVE  GEOMETRY         [XIX,  §  162 


The  required  planes  are  Q,  tangent  at  the  point  a,  and  T, 
tangent  at  the  point  b. 


Fig.  284. 


Problem  35.  To  pass  a  plane  tangent  to  a  cylinder  through  a 
given  point  without  the  surface.     (Two  results.) 

Analysis.  Through  the  given  point  pass  a  line  parallel  to 
the  elements  of  the  cylinder.  Since  the  recpiired  tangent 
plane  contains  some  element  of  the  surface,  this  line  must  lie 
in  the  required  tangent  plane.  Find  the  point  in  which  this 
line  pierces  the  plane  of  the  base  of  the  cylinder.  From  this 
point  draw  a  line  tangent  to  the  base.  Pass  the  required  tan- 
gent plane  through  the  tangent  line  and  the  line  first  drawn. 
In  general,  two  tangent  lines,  and  hence  two  tangent  planes, 
are  possible. 

Construction.  Case  I.  The  base  of  the  cylinder  lies  in  H  or 
V  (Fig.  285).  The  base  of  the  given  cylinder  lies  in  H.  The 
given  point  is  a.     Through  a  draw  the  line  B  (Bh,  B")  parallel 


XIX,  §  162] 


TANGENT  PLANES 

HT 


197 


to  the  elements  of  the  cylinder.  From  this  on  the  construc- 
tion is  the  same  as  for  the  corresponding  case  of  the  cone 
(Prob.  32,  Case  I).     The  required  planes  are  Q  and  T. 

Case  II.  Tlie  base  of  the  cylinder  does  not  lie  in  II  or  V 
(Fig.  286).  The  base  of  the  given  cylinder  lies  in  the  plane  Z, 
perpendicular  to  II.     Through  the  given  point  a,  draw  the 


Fig.  286. 


198 


DESCRIPTIVE  GEOMETRY         [XIX,  §  162 


line  B  parallel  to  the  elements  of  the  cylinder.  Then  proceed 
as  in  the  corresponding  case  of  the  cone  (Prob.  32,  Case  II). 
The  resulting  planes  are  Q  and  T. 

Problem  36.  To  pass  a  plane  tangent  to  a  cylinder  parallel  to  a 
given  line.     (Two  results.) 

Analysis.  No  point  of  the  required  tangent  plane  is  known 
in  advance,  as  in  the  corresponding  problem  with  the  cone 
(Prob.  33)  ;  hence  the  result  must  be  accomplished  by  indirect 
methods.  Through  the  given  line,  pass  a  plane  parallel  to  the 
elements  of  the  cylinder.  This  is  possible,  since  all  the 
elements   are  parallel.      This  plane  will  then  be  parallel  to 


Fig.  287. 


the  required  tangent  planes,  and  may  be  moved,  parallel  to 
itself,  until  tangent  to  the  cylinder.  The  manner  of  accom- 
plishing this  is  best  shown  in  the  construction. 

Construction.  Case  I.  Tlie  base  of  the  cylinder  lies  in  H  or 
V  (Fig.  287).  The  base  of  the  given  cylinder  lies  in  V. 
The  given  line  is  A.  Through  A  pass  the  plane  X  parallel  to 
the  elements  of  the  cylinder.  To  do  this,  assume  any  point,  c, 
in  A ;  through  c  draw  the  line  B,  parallel  to  the  cylinder ; 
pass  the  plane  X  through  the  lines  A  and  B.  (See  Prob.  7, 
§  107.)     Since  the  base  of  the  cylinder  lies  in  V,  VQ  is  now 


XIX,  §  162] 


TANGENT  PLANES 


199 


drawn  tangent  to  the  base  parallel  to  VX;  HQ  is  then  drawn 
parallel  to  HX.  The  traces  VT  and  HT  of  the  second  required 
tangent  plane  are  obtained  in  a  similar  manner. 

Case  II.  The  base  of  the  cylinder  lies  in  P  (Fig.  288). 
The  base  of  the  given  cylinder  is  a  circle  whose  center  is  o  ; 
find  the  P-projection  of  this  circle.  As  in  Case  I,  through  the 
given  line  A  pass  the  plane  X,  parallel  to  the  elements  of  the 
cylinder.  Since  the  base  of  the  cylinder  lies  in  P,  find  the 
profile  trace,  PX,  of  this  plane.  Draw  PQ  and  PT  parallel  to 
PX,  tangent  to  the  base  of  the  cylinder ;  these  are  the  profile 


Fig.  288. 


traces  of  the  required  tangent  planes.  From  PQ  and  PT  are 
obtained  the  //-and  F-traces  of  the  plain's,  HQ  and  ///'being 
parallel  to  HX,  while  VQ  and  FT7  are  parallel  to  VX. 

Cask  III.  The  hose  of  the  cylinder  tines  not  lit  in  //,  V,  or  P 
(Fig.  289).  The  base,  of  the  given  cylinder  lies  in  the  plane 
Zjj  perpendicular  to  V.    The  given  line  is  A.     Through  this 

line  is  passed,  as  in  the  previous  cases,  the  plane  X.  parallel  to 

the  elements  of  the  cylinder.  The  next  step  is  to  draw  a  tine, 
J",  which  shall  be  tangent  to  the  base  of  the  cylinder  and  at 


200 


DESCRIPTIVE  GEOMETRY         [XIX,  §  162 


the  same  time  parallel  to  the  plane  X.  This  line  must  lie  in 
the  plane,  Z,  of  the  base  of.  the  cylinder  (§  153)  ;  it  must  also 
be  parallel  to  some  line  in  the  plane  X  (§  104)  ;  hence  J  will 
be  parallel  to  the  line  of  intersection  of  the  planes  X  and  Z. 
Find  this  line  of  intersection,  L  (Prob.  12,  §  118).  Draw  the 
line  J  tangent  to  the  base  of  the  cylinder  parallel  to  the  line 


Fig.  289. 

L.     Through  J  pass  the  required  tangent  plane,  Q,  parallel 
toX 

A  second  result,  plane  T,  is  found  by  means  of  the  tangent 
line  K,  also  parallel  to  L.  Here  VT  is  not  located  either  by 
the  F-trace  of  K  or  the  point  where  HT  intersects  GL.  The 
construction  used  is  to  assume  some  point,  as  e,  on  K.  Through 
e  draw  the  line  M,  parallel  to  the  elements  of  the  cylinder. 
Then  the  line  M  lies  in  the  plane  T,  and  the  F-trace  t  is  a 
point  on  VT. 


CHAPTER   XX 


TANGENT   PLANES   TO   DOUBLE  CURVED  SURFACES  OF 
REVOLUTION 

163.  Double  Curved  Surfaces  of  Revolution.  A  double  curved 
surface  of  revolution  is  a  surface  formed  by  the  revolution  of 
any  curve  about  any  straight  line  as  an  axis,  provided  the 
resulting  surface  is  such  that  no  straight  lines  can  be  drawn 
on  it. 

Familiar  solids  whose  surfaces  are  double  curved  surfaces 
of  revolution  are  the  sphere,  the  various  ellipsoids,  and  the 
torus  (§  26). 

164.  Representation  of  Double  Curved  Surfaces  of  Revolution. 
In  solving  problems  involving  double  curved  surfaces  of  revo- 
lution, we  shall  place  the  axis  of  the  sur- 
face perpendicular  to  one  of  the  coordinate 
planes.  If  not  so  given,  the  transforma- 
tion can  be  effected  by  new  planes  of  pro- 
jection. 

In  Fig.  290  is  shown  a  general  case,  a 
vase  whose  outer  surface  is  a  double  curved 
surface  of  revolution.  The  axis  is  placed 
perpendicular  to  //. 

165.  Meridians.  The  F-projection,  Fig. 
290,  shows  the  true  shape  of  the  section 
made  by  a  plane  containing  the  axis  of  the 
surface.  Such  a  section  is  called  a  meridian 
section,  or  simply  a  meridian.  Any  plane 
which  contains  the  axis  is  called  a  meridian 

plane.      All  meridians  are   alike,  and   the  surface   is   usually 
considered  as  formed  by  the  revolution  of  its  meridian  section. 

201 


Fhi.  290. 


202 


DESCRIPTIVE  GEOMETRY  PCX,  §  165 


Fia.  290  (repeated). 


The  meridian  which  forms  the  outline  of  a  projection  of  the 
surface,  in  Fig.  290  the  F~-projection,  is  called  the  principal 
meridian.  The  plane  which  contains  this 
meridian  is  called  the  principal  meridian 
plane.  The  principal  meridian  plane  is  al- 
ways parallel  to  one  of  the  coordinate 
planes  ;  in  Fig.  290  it  is  parallel  to  V. 

166.  Parallels.  Let  a  double  curved  sur- 
face be  formed  by  the  revolution  of  its 
meridian  section  about  the  axis.  Each 
point  of  the  generating  meridian  describes 
a  circle  lying  in  a  plane  perpendicular  to 
the  axis  (§  74).  These  circles  are  called 
parallels  of  the  surface. 

The  projection  of  a  double  curved  sur- 
face of  revolution  on  a  plane  perpendicular 
to  its  axis  will  consist  of  one  or  more  cir- 
cles, which  are  the  projections  of  particular  parallels  of  the 
surface  (§§  75,  84).     For  example,  see  the  .ff-projection  of  the 
vase,  Fig.  290. 

167.  Projections  of  a  Point  in  a  Double  Curved  Surface  of  Revo- 
lution. Through  every  point  of  a  double  curved  surface  of 
revolution  a  circle  which  is  a  parallel  of  the  surface  (§  166)  can 
be  drawn.  A  point  will  lie  in  the  surface  if  its  projections  lie 
in  the  corresponding  projections  of  a  parallel  of  the  surface. 
With  the  axis  of  the  surface  perpendicular  to  either  H  or  V, 
one  projection  of  the  parallel  will  be  a  circle,  the  other  projec- 
tion a  straight  line  (§  75). 

For  example,  consider  the  ellipsoid  shown  in  Fig.  291,  where 
the  axis  of  the  ellipsoid  is  perpendicular  to  H.  Let  ah  be  a 
given  projection  of  a  point  in  the  surface.  Through  ah  draw 
the  iaT-projection,  a  circle,  Rh,  of  the  parallel  passing  through  a. 
Find  the  F-projection  of  the  circle  B,  namely,  the  straight  line 
Mv.  Project  from  ah  to  av  in  Rv ;  then  point  a(ah,  a")  lies  in 
the  surface.  The  given  ^-projection  may  also  represent  a 
point  b,  lying  on  the  symmetric  parallel  S  (Sh,  Sv). 


XX,  §  168]        SURFACES  OF  REVOLUTION 


203 


Let  C  be  given.  This  point  lies  on  the  parallel  T,  whose 
F-projection,  T",  is  a  straight  line  through  c".  The  //-projec- 
tion of  this  parallel  is  the  circle  T*.  In  Th  is  found  ch  by  pro- 
jection from  cv.  There  is  also  a  second  result ;  namely,  the 
point  d  (d",  dh) 

If  a  surface  is  wholly  convex,  as  in  this  example,  any  point 
assumed  in  either  projection  represents  two  points  in  the  sur- 


FlG.  291. 


face,  unless  the  assumed  projection  lies  in  the  contour  (outline) 
of  the  surface.  Thus,  if  e"  is  assumed  in  the  ^projection,  3f", 
of  the  principal  meridian  (§  105),  the  single  //-projection,  eh, 
lies  in  the  straight  line  Mh>  which  is  the  //-projection  of  the 
principal  meridian.  If  /*  be  assumed  in  the  H"-projection,  /"', 
of  the  greatest  parallel  (§  1  <><>), /"  lies  in  the  straight  line  P  . 
the  ^projection  of  this  parallel. 

168.    The  Sphere.     The  sphere  possesses  the  unique  property 
that  every  plane  section  is  a  circle.     All  planes  which  contain 


204 


DESCRIPTIVE  GEOMETRY  [XX,  §  168 


the  center  of  the  sphere  intersect  the  surface  in  circles  of  the 
same  size,  known  as  great  circles.  Any  diameter  of  the  sphere 
may  be  taken  as  its  axis.  In  the  solution  of  problems  involv- 
ing the  sphere,  advantage  is  usually  taken  of  some  or  all  of 
these  properties.  The  solution  thus  becomes  a  particular  solu- 
tion, and  the  sphere,  in  consequence,  is  not  a  good  surface  to 
use  to  illustrate  the  general  case  of  a  double  curved  surface  of 
revolution. 

In  the  problems  of  tangencies  which  follow,  we  shall  treat 
the  sphere  as  a  particular  case,  apart  from  the  general  problem. 


Fig.  292. 


Fig.  293. 


On  account  of  the  perfect  symmetry  of  the  sphere,  and  the 
possibility  of  taking  any  diameter  as  an  axis,  students  often 
have  more  difficulty  in  visualizing  the  relation  of  points  on  the 
surface  than  with  a  less  symmetric  surface.  This  difficulty 
usually  disappears  by  visualizing  first  a  hemisphere,  as  shown 
in  Fig.  292.  This  figure  represents  the  front  half  of  a  sphere, 
bounded  by  the  great  circle  0(0",  Ch),  and  the  portion  of  the 
surface  represented  by  each  projection  appears  more  clearly 
than  when  the  entire  sphere  is  given,  as  in  Fig.  293. 

169.  The  Torus.  The  surface  of  a  torus  (Fig.  294)  will 
usually  be  taken  to  illustrate  the  general  case  of  a  surface  of 
revolution.     This  surface  is  divided  by  the  circles  C  and  D 


XX,  §  170]         SURFACES  OF  REVOLUTION 


205 


into  two  portions  :  the  so-called  outer  portion,  doubly  convex, 
where  a  tangent  plane  contains  but  a  single  point  of  the  sur- 
face, and  an  inner  portion,  concavo-convex,  where  a  tangent 
plane  at  any  point  also  intersects  the  surface. 

Certain  properties  of  this  surface  have  already  been  given. 
Thus  in  §  86,  it  has  been  shown  that  any  plane  perpendicular 
to  the  axis  between  C  and  D  cuts  the  surface  in  two  circles  A 


Fig.  294. 


and  B,  which  are  parallels  of  the  surface.  Hence,  if  the  sur- 
face is  given  as  in  Fig.  294,  points  assumed  in  the  F-projection 
may  represent,  according  to  their  position,  one,  two,  three,  or 
four  points  lying  in  the  surface.  A  point  assumed  in  the  H- 
projection,  however,  as  in  the  ellipsoid,  Fig.  291,  can  never 
represent  more  than  two  points  in  the  surface. 

170.    Tangent  Planes  to  Double  Curved  Surfaces  of  Revolution, 
(a)  A  plane  tangenl  to  a  double  curved  surface  of  revolution 

is  tangent,  in  general,  at  but  a  single  point.  Exceptions  occur, 
however;  for  example,  in  the  torus,  Fig.  294,  the  planes  which 
contain  the  circles  C  and  D  are  tangent  at  every  point  in 
these  circles. 


206 


DESCRIPTIVE  GEOMETRY 


[XX,  §  170 


(6)  Through  every  point  in  the  surface  two  curves  of  known 
properties  can  be  drawn,  namely,  the  meridian  (§  165)  and 
the  parallel  (§  166)  which  pass  through  the  given  point. 

(c)  Planes  which  are  tangent  to  points  lying  in  the  same 
parallel  of  the  surface  intersect  the  axis  at  the  same  point. 

(d)  The  normal  (§  156)  at  any  point  of  the  surface  lies  in 
the  meridian  plane  passing  through  the  point. 

(e)  Every  normal  to  the  surface  intersects  the  axis. 

(/)  Normals  to  points  lying  in  the  same  parallel  of  the  sur- 
face intersect  the  axis  at  the  same  point. 


Fig.  295. 


Problem  37.  To  pass  a  plane  tangent  to  a  sphere  at  a  given 
point  in  the  surface. 

Analysis.  Through  the  given  point,  draw  two  circles  lying 
in  the  surface.  Draw  tangent  lines,  one  to  each  circle,  at  the 
given  point.  Find  the  plane  determined  by  the  two  tangent 
lines.     (See  §  155.) 

Construction  (Fig.  295).  Let  a  (ah,  av)  be  the  given  point. 
Through  a  we  can  draw  two  circles  which  have  simple  projec- 


XX,  §  170]         SURFACES  OF  REVOLUTION  207 

tions,  namely  the  circle  C  (CA,  Cv)  parallel  to  Fand  the  circle 
I)  (Dh,  Dv)  parallel  to  H.  Through  a  draw  the  Hue  J  (Jh,  J") 
tangent  to  the  circle  C,  and  line  K  (A"  \  Kv)  tangent  to  the 
circle  D  (§  152).  The  required  tangent  plane,  T,  is  the  plane 
determined  by  the  lines  J  and  A"(§§  102,  106). 

A  second  result  is  the  plane  Q  which  is  tangent  at  the 
point  b  (bh,  bv)  and  obtained  in  a  similar  manner. 

Problem  38.  To  pass  a  plane  tangent  to  a  double  curved  surface 
of  revolution  at  a  given  point  in  the  surface. 

First  Analysis.  The  meridian  and  the  parallel  which  pass 
through  the  given  point  are  two  known  curves  of  the  surface 
(§  170,  b).  At  the  given  point,  draw  tangent  lines,  one  to 
each  of  these  curves.  Find  the  plane  determined  by  the  two 
tangent  lines  (§  155). 

Second  Analysis.  Draw  the  normal  to  the  surface  at  the 
given  point.  Pass  the  required  tangent  plane  through  the 
point  perpendicular  to  the  normal  (§  156). 

Third  Analysis.  Revolve  the  surface  about  its  own  axis  until 
the  given  point  is  in  the  principal  meridian  plane.  In  this 
position  the  tangent  plane  will  show  edgewise  as  a  line  tan- 
gent to  the  outline  of  the  surface  at  the  given  point.  Counter- 
revolve  the  surface  together  with  the  tangent  plane  to  the 
original  position. 

The  actual  construction,  if  made  by  the  first  analysis,  must 
often  be  supplemented  by  the  use  of  auxiliary  lines  to  obtain 
points  within  the  limits  of  the  drawing.  It  has  also  been 
shown,  in  §  156,  that  while  the  normal  alone  determines  the 
plane  in  space,  auxiliary  lines  must  be  used  to  locate  the 
traces  in  the  drawing.  P>ut  by  the  third  analysis,  it  is  possible 
in  nearly  every  case  to  determine  one  trace  of  the  tangent 
plane  neatly  and  rapidly.  We  shall  therefore  begin  the 
solution  by  the  third  analysis;  then,  to  find  the  second  trace 
of  the  plane,  introduce  as  auxiliaries  one  or  nunc  of  the 
lines  indicated  by  the  first  and  second  analyses. 

Construction  (Fig.  296).  The  surface  chosen  to  illustrate 
this,  the  general  case,  is  that  of  a  torus.     Let  a  (a*,  a ")  be  the 


208 


DESCRIPTIVE  GEOMETRY 


[XX,  §  170 


given  point  in  the  surface.  Revolve  the  torus  about  its  axis 
until  a  is  in  the  principal  meridian  plane  Y.  The  path  of  a 
will  be  the  parallel  C,  and  the  resulting  position  ar  (arh,  arv). 
At  arv  we  can  draw  by  inspection  the  edge  view  VTX  of  the 


V                  '   Dv 

\  \K 

|/fc 

Lv 

y 

\^       s 

^  \> 

xk 

,/t. 

X 

Fig.  296. 

tangent  plane  at  this  point.  The  if-trace  of  this  plane  is  HT^. 
Counter-revolving  the  torus  to  the  first  position,  HTX  takes 
the  final  position  HT  (reverse  of  §  134).  The  tangent  plane, 
as  shown  at  VT1}  intersects  the  axis  at  the  point  /  (/",  fh), 
which  becomes  a  fixed  point  for  all  planes  tangent  at  any  point 
in  the  parallel  C  (§  170,  c).     Hence,  connecting  /  and  a,  we 


XX,  §  170]         SURFACES  OF  REVOLUTION  209 

have  the  line  K  (Kv,  Kh),  evidently  tangent  to  the  meridian  E 
and  lying  in  the  tangent  plane  at  a.  The  vertical  trace,  VT, 
of  the  required  tangent  plane  is  now  determined  by  the  V- 
trace,  ^  of  K,  together  with  the  point  in  which  HT  intersects 
GL.  In  this  case  the  line  «/,  tangent  to  the  parallel  C  at  a, 
might  be  used  instead  of  the  line  K. 

Let  b  (bh,  b"),  Fig.  296,  be  a  second  point.  Let  b  be  revolved 
into  the  principal  meridian  at  br  (brh,  br").  At  brv  we  may  draw 
by  inspection  the  edge  view  VQX  of  the  plane  tangent  at  this 
point.  This  edge  view  does  not  intersect  the  axis  within 
reach,  as  was  the  case  with  point  a;  but  the  plane  Qt  inter- 
sects H  in  the  ff-trace  HQX.  AVe  can  therefore  obtain  HQ  by 
revolving  HQX  about  the  axis  of  the  torus,  as  was  done  for 
point  a.  The  line  L,  tangent  to  D  at  b,  is  one  line  in  the 
required  tangent  plane  Q.  This  gives  one  point,  ti}  in  VQ. 
But  since  HQ  does  not  intersect  GL  within  the  figure,  VQ  is 
still  undetermined.  Let  us  try  the  normal.  Draw  first  the 
normal,  NT,  at  the  point  brv.  This  normal  intersects  the  axis 
at  the  point  gv.  Then  the  normal  at  the  point  b  passes  through 
#(§  170,/).  Connecting^  and  bv  gives  the  F-projection,  N". 
Since  the  plane  Q  is  perpendicular  to  the  normal  N,  VQ  is 
now  drawn  through  ?4  perpendicular  to  N"  (§  11-5). 

Where  is  the  //-projection  of  the  normal  A".'  Note  that 
'this  projection  would  not  be  needed  in  any  event,  since  it 
would  merely  give  the  direction  of   HQ. 

A  second  example,  involving  some  additional  features  of 
construction,  is  given  in  Fig.  297.  The  given  point  is  so 
chosen  in  the  F-projectiou  thai  it  represents  lour  points  in 
the  surface  (§  169).  Two  tangent  planes  are  shown,  one  at  an 
outside  point,  ami  one  at  an  inside  point.  Tin-  plane  V  is  tan- 
gent at  point  a.  This  point  lies  in  the  parallel  /«,'  I  /''.  /  >.  ■> 
that  it  revolves  into  t be  principal  meridian  plane  at  n  (ark,  a  I. 
The  trace  HQ  is  obtained  by  revolving  HQX  about  the  axis  of 
the  torus,  as  in  Fig.  296.  The  meridian  tangent  A',  and  the 
line.  J  tangent  to  the  parallel  h\  give  tin'  P-traces  /,  and  /.., 
thus  locating  VQ.  The  normal,  N,  to  the  plane  Q,  although 
not  needed,  is  shown  in  the  F-projection ;  if  here  furnishes  a 


210  DESCRIPTIVE  GEOMETRY  [XX,  §  170 

check  on  the  construction,  since  Nv  should  be  perpendicular  to 
VQ(%  112). 

The  plane  T,  Fig.  297,  is  tangent  at  the  point  c.  This  point 
lies  in  the  parallel  F  (Fh,  F")  ;  hence  it  revolves  into  the  prin- 
cipal meridian  at  cr  (crh,  crv).  At  crv  draw  the  edge  view,  VTU 
of  the  tangent  plane,  also  the  normal,  UT,  to  this  plane.  Then 
HTX  revolves  to  HT.  Since  VTX  intersects  the  axis  at  3",  the 
line  M  (Mv,  3Ih)  is  a  line  in  the  required  tangent  plane  at  c. 
But  the  line  M  gives  no  accessible  point  on  VT;  neither  does 
the  line  L,  drawn  tangent  to  the  circle  F  at  point  c.  Draw 
the  P"-projection,  Uv,  of  the  normal  to  the  plane  T  through  4" 
and  C  (§  170,/).  Note  that  now  but  one  point  on  VT  is  nec- 
essary. Why  ?  Since  the  line  M  lies  in  the  plane  T,  let  the 
point  5  (5A,  5")  be  some  assumed  point  on  M.  Through  the 
point  5  draw  the  line  X  (Xk,  X")  parallel  to  the  line  L  (§  108, 
Ex.  1).  Find  the  F-trace,  t-0,  of  X.  Draw  the  recpiired  trace, 
VT,  through  t0  perpendicular  to  Uv. 

171.  Tangent  Planes  which  Contain  a  Given  Line.  A  plane 
tangent  to  a  double  curved  surface  of  revolution  may,  under 
certain  conditions,  be  determined  by  the  fact  that  the  plane 
must  contain  a  given  straight  line.  For  example,  with  sur- 
faces like  the  sphere  and  the  ellipsoids,  it  is  evident  that  the 
necessary  condition  is  that  the  line  shall  not  intersect  the  sur- 
face;  then,  in  general,  two  tangent  planes  are  jwssible.  But, 
whatever  the  form  of  the  surface,  if  a  solution  exists  in  space, 
the  traces  of  the  plane  can  be  found. 

We  are  not  prepared  to  take  up  at  this  time  the  general 
case  of  a  plane  through  any  line  tangent  to  any  double  curved 
surface  of  revolution.  The  solution  requires  the  use  as  an 
auxiliary  of  a  surface  whose  properties  have  not  yet  been  dis- 
cussed. Particular  solutions  can  be  found,  however,  for  the 
following  cases. 

(1)  The  given  surface  a  sphere,  the  line  general,  in  any 
position  not  intersecting  the  sphere. 

(2)  Any  double  curved  surface  of  revolution,  with  the  line 
in  particular  positions  with  respect  to  the  axis  of  the  surface. 


XX,  §  171]        SURFACES  OF  REVOLUTION 


211 


£ 


I    cn 


Fiu.  '.'97. 


212 


DESCRIPTIVE  GEOMETRY 


[XX,  §171 


Problem  39.  To  pass  a  plane  tangent  to  a  sphere  through  a 
given  line  without  the  surface. 

Analysis.  This  analysis  is  shown  pictorially  in  Fig.  298. 
Let  A  be  the  given  line,  and  o  the  center  of  the  given  sphere. 
Let  Q  and  T  represent  the  required  tangent  planes.  Pass  an 
auxiliary  plane,  X,  perpendicular  to  A,  through  the  center  of 
the  sphere.     This  plane  intersects  A  in  the  point  c.     Since  X 

contains  the  center 
of  the  sphere,  the  in- 
tersection with  the 
sphere  is  a  great  cir- 
cle. This  circle  evi- 
dently contains  the 
points  of  tangency  of 
the  planes  Q  and  T. 
From  c  draw  the  line 
E  tangent  to  the 
great  circle.  This 
can  be  done,  since  c 
and  the  circle  are  in 
the  same  plane,  X 
(§153).  Then  E  will 
lie  wholly  in  the  tan- 
gent plane  Q,  which 
is  now  determined, 
since  we  know  two  lines,  A  and  E,  lying  in  it.  The  plane  T  is 
similarly  determined  as  the  plane  which  contains  the  given 
line  A  and  the  tangent  line  F. 

An  alternative  method  for  determining  the  plane  T  is  also 
shown  in  Fig.  298.  From  o  draw  the  line  N  perpendicular  to  F. 
This  line  lies  in  the  plane  X,  and  passes  through  the  point  of 
tangency  of  F  and  the  sphere.  The  line  N  is  thus  the  normal 
to  the  tangent  plane  T.  Hence  T  may  be  determined  as  the 
plane  which  contains  the  given  line  A  and  is  perpendicular  to 
N.  We  may  observe  in  passing  that  it  is  not  possible,  in 
general,  to   pass   a  plane   through  one  line  perpendicular   to 


Fig.  298. 


XX,  §171]  SURFACES  OF  REVOLUTION 


213 


another  line,  but  the  relative  positions  of  the  lines  A  and  JV 
are  such  in  this  case  that  it  may  be  done. 

Construction  (Fig.  299).  Let  A  be  the  given  line,  and  o  the 
center  of  the  given  sphere.  Find  the  plane,  X,  which  con- 
tains o  and  is  perpendicular  to  A  (Prob.  10,  §  115).  Find  the 
point,  c,  in  which  A  intersects  X  (Prob.  13,  §  119).     Since  the 


Fm.  299. 

plane  X  is  oblique  to  both  //  and  V,  the  great  circle  in  which 
X  intersects  the  sphere  will  project  as  an  ellipse  in  each  new. 
To  avoid  drawing  these  ellipses,  revolve  the  plane  X  about 
one  of  its  traces  into  //  or  V  (§  L38).  Lei  X  be  revolved 
about  HX  into  H.  Tin;  point  c  revolves  bo  c,  (Prob.  21, 
Working  Rule,  §  138);  and  the  point  o  revolves  to  <>,.  The 
great  circle  lying  in  X  will  now  appear  in  true  shape  and  size, 


214 


DESCRIPTIVE  GEOMETRY 


[XX,  §  171 


and  may  be  drawn  at  once,  with  or  as  center.  From  cr  draw 
the  two  tangents,  Er  and  FT,  to  this  circle.  These  are  two 
lines  lying  in  the  plane  X  (see  the  Analysis),  and  their  pro- 
jections may  be  obtained  by  counter-revolving  the  plane  X 
(§  147).     The  tangent  Er  intersects  HX  at  s3*,  whence  s3v  is  on 


Fig.  299  (repeated). 


GL,  and  the  projections  Eh,  Ev  are  determined  by  the  fact 
that  the  line  E  passes  through  the  point  c  (reverse  of  Fig.  221, 
§  138).  Pass  the  required  tangent  plane  Q  through  the  lines 
E  and  A  (Prob.  6,  §  106). 

The  tangent  line  F  does  not  intersect  HX  within  reach.  In- 
stead of  attempting  to  find  the  projections  of  F,  let  us  deter- 
mine the  tangent  plane  by  means  of  the  normal.  Through  oT 
draw  Nr  perpendicular  to  FT.     The  line  N  lies  in  the  plane  X 


XX,  §  171]        SURFACES  OF  REVOLUTION 


215 


(see  the  Analysis) ;  hence  the  intersection  of  Nr  and  HX  is  the 
//-trace  of  N.  Using  this  trace  and  the  point  o,  determine  the 
projections,  JV*  and  Nv,  of  N.  The  required  tangent  plane  T 
contains  A  and  is  perpendicular  to  N.  Let  sx  and  tx  be  the 
traces  of  A.  Through  sx  draw  HT  perpendicular  to  JV* ; 
through  tx  draw  VT  perpendicular  to  N"  (§  112). 


Fig.  300. 

Special  Case.  Tlie  given  line  is  parallel  to  H  or  V.  Let  the 
given  line  A,  Fig.  300,  be  parallel  to  77.  Assume  a  secondary 
ground  line,  0\L\,  perpendicular  to  Ah.  The  line  A  will  then 
project  as  a  point,  A?  (§  70).  Project  the  center  of  the  sphere 
to  of,  and  draw  the  sphere.  The  edge  views,  VXQ  and  V\T,  of 
the  required  tangent  planes  may  now  lie  drawn  directly  through 
A{.  The  plane  Q  (HQ,  VQ)  requires  no  further  explanation. 
The  trace  HT  is  readily  obtained,  but  for  Fran  auxiliary  line 
is  necessary.  The  line  used  here  is  /;,  parallel  to  the  line  A. 
The  projection  /»',''.  a  point,  is  assumed  in  V\T.  From  />','  ;l"■ 
obtained  Bh and  Bv.     The F-trace, t2, of  B  locates  a  point  in  I"/'. 


216 


DESCRIPTIVE  GEOMETRY  [XX,  §  171 


Problem  40.     To  pass  a  plane  tangent  to  a  double  curved  surface 
of  revolution  through  a  given  line.     (Special  Cases.) 

Case  I.     The  given  line  intersects  the  axis  of  the  surface. 
Analysis.     Using  the  point  in  which  the  given  line  intersects 


Fig.  301. 


the  axis  as  a  vertex,  circumscribe  the  given  surface  by  a  cone 
of  revolution.  Pass  the  required  tangent  planes  through  the 
given  line  tangent  to  the  auxiliary  cone. 

Construction  (Fig.  301).  The  given  surface  is  an  oblate 
spheroid,  with  its  axis  perpendicular  to  H.  The  given  line  A 
intersects  the  axis  at  the  point  c  (c*,  c!).  Prom  c*  draw  tan- 
gents to  the  P"-projection  of  the  given  surface;  these  tangents 
represent  the  auxiliary  cone  in  the  ^projection.  The  base  of 
this  cone  on  H  is  the  circle  Bh.  The  if-traces,  HQ  and  HT,  of 
the  required  tangent  planes  pass  through  the  JY-trace,  su  of  A, 
and  are  tangent  to  the  circle  Bh.     (Compare  Prob.  32,  §  162.) 


XX,  §  171]         SURFACES  OF  REVOLUTION  217 

The  F-traces,  VQ  and  VT,  pass  through  the  U-traee,  tu  of  A. 
A  second  point  on  VT  is  here  obtained  by  finding  the  F-trace, 
t2,  of  the  element,  E,  hi  which  the  plane  T  is  tangent  to  the 
auxiliary  cone. 

Case  II.      TJie  given  line  is  parallel  to  the  axis  of  the  surface. 

Analysis.  The  axis  of  the  given  surface  being  placed,  as 
usual,  perpendicular  to  one  of  the  coordinate  planes,  for  ex- 
ample H,  the  .//-projection  of  the  given  line  will  be  a  point. 
The  ^-traces  of  the  tangent  planes  will  therefore  be  their  edge 
views,  and  the  planes  may  be  drawn  by  inspection.  We  pro- 
ceed similarly  if  the  axis  of  the  surface  is  perpendicular  to  V 
or  P. 

No  figure  to  illustrate  the  construction  is  deemed  necessary 

Case  III.     The  given  line  is  perpendicular  to,  but  does  not 
fntersect,  the  axis  of  the  given,,  surface. 

First  Analysis.  Let  the  axis  of  the  given  surface  be  per- 
pendicular to  H.  Then  the  given  line  is  parallel  to  H.  Re- 
volve the  line  about  the  axis  of  the  surface  until  the  line  is 
perpendicular  to  V,  and  projects  as  a  point  thereon.  The  edge 
views  of  tangent  planes  may  now  be  drawn  by  inspection. 
Revolve  these  planes  about  the  axis  of  the  surface  until  they 
contain  the  original  position  of  the  given  line.  We  proceed 
similarly  if  the  axis  of  the  surface  is  perpendicular  to  For  P. 

Second  Analysis.  The  given  line  may  be  projected  as  a  point 
by  choosing  a  secondary  plane  of  projection  perpendicular  to 
the  line.  This  plane  necessarily  will  be  parallel  to  the  axis  of 
the  given  surface.  Hence  the  surface  may  be  projected  readily, 
and  the  edge  views  of  the  tangent  planes  may  be  drawn  by 
inspection.  The  traces  of  the  planes  can  then  be  obtained 
from  the  edge  views. 

Since  the  met  hod  of  the  second  analysis  aecessitates  the  '-1111 
struction  of  an  additional  projection  oJ  the  given  surface,  if  is 
usually  easier  to  employ  the  firsl  analysis.     The  following  con- 
struction is  made  by  the  first  analysis. 

Construction  (Fig.  302).  The  given  surface  is  thai  of  a  aolid 
torus,  only  the  outer  (doubly-convex)   portion  of  the  curved 


218 


DESCRIPTIVE  GEOMETRY 


[XX,  §  171 


surface  being  retained.  The  axis  of  the  surface  is  perpen- 
dicular to  V.  The  given  line  A  (Av,  Ah)  is  parallel  to  V.  Re- 
volve A  about  the  axis  of  the  torus  to  the  position  Ax  (Af, 
Axh),  where  the  ^-projection,  Axh,  is  a   point.     Through  Ax 


Fig.  302. 


draw  the  tangent  planes  &  (HQ1}  VQi)  and  Tx  (HTU  VTJ.  Re- 
volve Qi  and  Tx  about  the  axis  of  the  torus  to  the  positions 
Q  (VQ,  HQ)  and  T  ( VT,  HT),  so  that  each  plane  contains  the 
line  A  (§  134).  Then  Q  and  T  are  the  required  tangent 
planes. 


CHAPTER   XXI 

THE  INTERSECTION  OF  CURVED  SURFACES  BY  PLANES 

172.  Classification  of  Curved  Surfaces.  Curved  surfaces  may 
be  divided  into  three  classes  : 

(a)  Ruled  surfaces,  on  which  a  straight  line,  or  rectilinear  ele- 
ment, can  be  drawn  through  any  point  of  the  surface ; 

(b)  Surfaces  of  revolution,  on  which  a  circle,  lying  in  a  plane 
perpendicular  to  the  axis,  can  be  drawn  through  every  point 
of  the  surface  ; 

(c)  All  other  curved  surfaces,  on  which,  in  general,  neither 
straight  lines  nor  circles  can  be  drawn. 

Certain  surfaces,  for  example  the  cone  and  cylinder  of 
revolution,  belong  to  both  classes  (a)  and  (b). 

173.  The  Intersection  of  a  Ruled  Surface  and  a  Plane.  Select  a 
sufficient  number  of  the  rectilinear  elements  of  the  surface. 
Find  the  points  in  which  these  straight  lines  intersect  the 
given  plane.     These  points  lie  in  the  required  intersection. 

In  the  process  of  finding  where  a  straight  line  intersects  a 
plane,  it  is  usual  to  puss  through  the  line  an  auxiliary  plane 
perpendicular  to  //  or  V.  (See  Prob.  13,  §  119.)  Hence  the 
elements  of  the  ruled  surface  are  often  chosen  by  passing 
through  the  surface  a  series  of  planes  perpendicular  to  //  or 
V,  these  same  planes  being  then  used  as  auxiliaries  for 
finding  the  points  in  which  the  resulting  elements  intersect 
the  secant  plane. 

174.  The  Intersection  of  a  Surface  of  Revolution  and  a  Plane. 
An  auxiliary  plane  taken  perpendicular  to  the  axis  of  the 
given  surface  will  cut  one  or  more  circles  from  the  surface, 

and,  in  general,  a  straight  line  from  the  given  pli Points 

of  intersection,  if  any,  of  the  straighl  line  with  the  circle 
or  circles,  will  lie  in  the  required  intersection  of  the  plane  and 
the  surface. 

219 


220  DESCRIPTIVE  GEOMETRY         [XXI,  §  174 

In  order  that  the  circular  sections  of  the  given  surface  may 
be  projected  readily,  the  axis  of  the  surface  should  be  perpen- 
dicular to  one  of  the  coordinate  planes.  (Compare  §§  84,  86, 
164.) 

175.  The  Intersection  of  Any  Curved  Surface  and  a  Plane.  In 
general,  the  intersection  of  a  surface  and  a  plane  can  be  found 
by  drawing  on  the  given  surface  any  series  of  curves  which 
are  known  to  lie  in  the  surface,  and  then  finding  the  points  in 
which  these  curved  lines  intersect  the  given  plane. 

If  the  curves  which  can  be  drawn  in  the  surface  are  plane 
curves,  the  planes  of  these  curves  may  be  used  as  auxiliary 
planes,  and  the  process  of  finding  the  intersection  is  similar  to 
that  for  a  surface  of  revolution. 

176.  Method  by  Means  of  Secondary  Projections.  Any  plane 
may  be  seen  edgewise  by  projecting  on  a  suitable  secondary 
plane  of  projection  (§  70).  A  general  method  of  finding  the 
intersection  of  any  surface  or  solid  with  a  plane  is,  therefore, 
to  make  a  secondary  projection  of  the  surface  and  the  secant 
plane  so  that  the  plane  appears  edgewise.  This  reduces  one 
projection  of  the  required  intersection  to  a  straight  line. 
Since  this  method  involves  the  construction  of  a  new  pro- 
jection of  the  given  surface,  as  well  as  of  the  secant  plane,  it 
will  be  found  of  doubtful  advantage  in  the  case  of  surfaces 
consisting  of  rectilinear  elements.  The  method  is,  however,  a 
good  one  with  surfaces  of  revolution,  whose  projections,  from 
any  point  of  view  perpendicular  to  the  axis,  are  alike. 

177.  Visibility  of  the  Curve  of  Intersection.  In  finding  the 
intersection  of  a  plane  with  a  curved  surface  or  solid,  the  plane 
will  be  considered  to  be  transparent,  and  in  general  no  portion 
of  the  surface  or  solid  will  be  removed.  The  various  portions 
of  the  resulting  line  of  intersection,  therefore,  will  be  visible 
or  invisible,  according  as  the}^  lie  in  a  visible  or  an  invisible 
portion  of  the  surface.     (Compare  §  85.) 

178.  A  Rectilinear  Tangent  to  the  Curve  of  Intersection.  At 
any  point  in  the  intersection  of  any  curved  surface  and  a 


XXI,  §  179]     CURVED  SURFACES  AND  PLANES  221 

plane,  a  rectilinear  tangent  to  the  intersection  may  be  drawn 
without  knowing  or  considering  the  properties  of  the  curve  of 
intersection.  For  the  tangent  line  must  lie  in  each  of  two 
planes  :  first,  in  the  given  secant  plane,  since  every  tangent  to 
a  plane  curve  lies  in  the  plane  of  the  curve  (§  153) ;  and 
second,  hi  the  plane  tangent  to  the  given  surface  at  the 
point  in  question  (§  154).  Hence,  if  the  plane  tangent  to  the 
surface  at  the  given  point  can  be  found,  the  line  of  intersection 
of  this  plane  with  the  given  secant  plane  will  be  tangent  to 
the  curve  of  intersection  at  the  given  point. 

179.  Development.  A  curved  surface  is  developable  when  it 
can  be  unrolled  into  a  plane,  without  extension,  compression, 
or  distortion  of  any  kind.  Only  surfaces  which  consist  of 
rectilinear  elements  are  developable ;  and  of  these  surfaces, 
only  those  forms  in  which  every  two  consecutive  elements  lie 
in  the  same  plane,  that  is,  either  intersect  or  are  parallel. 
Cones  and  cylinders  are  developable,  and  are  the  only  forms 
of  developable  curved  surfaces  commonly  used  in  practical 
work. 


CHAPTER   XXII 

INTERSECTION  OF  PLANES  WITH  CONES  AND 
CYLINDERS 

180.  The  Intersection  of  a  Cone  or  Cylinder  with  a  Plane.  The 
cone  and  the  cylinder  are  ruled  surfaces.  Hence  the  method 
of  finding  points  in  the  intersection  is  that  described  in  §  173. 

In  the  three  problems  which  follow,  in  addition  to  finding 
the  curve  of  intersection  we  shall  draw  a  tangent  line  to  some 
point  of  the  intersection  (§  178),  and  develop  the  curved  sur- 
face, showing  the  section  and  the  tangent  line. 

Problem  41 .     To  find  the  intersection  of  a  cone  and  plane. 

Case  I.     The  base  of  the  cone  lies  in  H  or  V. 

A.    The  Intersection.     Analysis.     See  §  173. 

Construction  (Fig.  303).  The  base  of  the  given  cone  lies  in 
H.  The  given  secant  plane  is  Q.  Eight  elements  have  been 
chosen  on  the  cone,  distinguished  by  the  numbers  1  to  8  around 
the  base.  Find  the  points  in  which  these  elements  intersect 
the  plane  Q  by  passing  through  them  auxiliary  planes  perpen- 
dicular to  H  or  V  (Prob.  13,  Usual  Method,  §  119).  The 
planes  here  chosen  are  the  planes  X,  Y .  .  .  X,  perpendicular 
to  H.  The  intersection  of  planes  X  and  Q  is  the  line  A. 
The  intersection  of  Av  and  the  F-projection  of  the  element 
0-3  locates  13",  one  point  of  the  required  intersection.  The 
intersection  of  planes  Z  and  Q  is  the  line  C,  the  projection  C 
being  parallel  to  VQ.  Since  plane  Z  contains  two  elements  of 
the  cone,  the  construction  locates  two  points,  11  on  element 
0-1  and  15  on  element  0-5.  But  the  planes  Y,  M,  and  X  are 
so  situated  that  in  attempting  to  find  their  lines  of  intersection 
with  Q,  only  one  point  in  each  line  can  be  found  within  the 
limits  of  the  figure,  namely,  the  points  b",  d",  and  e"  on  the 
ground  line. 

222 


XXII,  §  180] 


PLANE  AND  CONE 


223 


B.   The  Common  Point.     Since  the  planes  X,  T,  Z,  M,  and 

N  are  perpendicular  to  H,  it  follows  that  the  lines  of  inter- 


im [Q,  303 


section  A,  B,  C,  etc.,  will  have  their  H"-projections  coincident 
with  HX,  II Y,  IIZ,  etc.,  respectively.     But  the  lines  of  inter- 


224  DESCRIPTIVE  GEOMETRY        [XXII,  §  180 

section  A,  B,  C,  etc.,  are  all  lines  in  one  plane,  namely,  the 
plane  Q.  Now  if  a  number  of  lines  known  to  be  in  the  same 
plane  appear  to  intersect  in  a  common  point  in  one  view,  it 
can  only  be  because  the  lines  actually  intersect  in  space. 
Hence  in  the  other  view  the  lines  must  also  pass  through  a 
common  point.  In  the  figure,  the  iZ-projection  of  this  point 
is  nh,  while  nv  lies  on  A".  The  other  F'-projections,  Bv,  D", 
and  Ev,  may  now  be  drawn  through  nv  and  the  points  bv,  dv, 
and  ev  on  GL,  already  determined.  The  intersections  of  Bv, 
Dv,  and  Ev  with  the  elements  lying  in  the  corresponding 
auxiliary  planes  locate  the  remaining  points  in  the  curve  of 
intersection. 

C.  Visibility  of  the  Curve  of  Intersection.  (See 
§  177.)  In  the  jff-projection,  the  point  11,  on  the  highest 
element,  is  evidently  visible.  This  determines  the  visible  side 
of  the  curve,  which  will  become  invisible  at  the  points  13 
and  17,  which  lie  on  the  boundaries  of  the  visible  surface  of 
the  cone.  Similarly,  in  the  F"-projection,  the  point  17,  on  the 
extreme  front  element,  is  visible.  The  visible  side  of  the 
curve,  beginning  at  11,  passes  through  17  and  ends  at  15. 

D.  A  Line  Tangent  at  a  Given  Point  in  the  Curve 
of  Intersection.     Analysis.     See  §  178. 

Construction.  Let  12,  lying  on  the  element  0-2,  be  the 
given  point.  Find  the  plane  S(HS,  VS),  which  is  tangent  to 
the  cone  at  the  point  12  (Prob.  31,  §  162).  Find  the  intersec- 
tion of  S  with  the  given  plane  Q  (Prob.  12,  §  118).  This  line 
of  intersection  T (Th,  Tv)  is  the  required  tangent  line.  As  a 
check,  it  should  pass  through  the  point  12,  and  show  tangent 
to  the  curve  of  intersection  in  each  projection  (§  152). 

E.  Development  of  the  Curved  Surface  of  the  Cone. 
Analysis.  The  entire  curved  surface,  between  the  vertex  and 
the  base,  should  be  developed  first.  Any  other  line  drawn  on 
the  surface  can  then  be  put  in.  The  conical  surface  is  divided 
by  its  elements  into  pieces  bounded  by  two  straight  lines  and 
a  curve.  These  pieces  are  approximately  plane  triangles,  the 
approximation  being  closer  the  nearer  the  elements  are  to- 
gether.    The  curved  surface  may  then  be  developed  as  if  the 


XXII,  §  180] 


PLANE  AND  CONE 


225 


cone  were  a  many-sided  pyramid,  the  triangular  faces  being 
built  up  from  the  true  lengths  of  their  sides.     (See  §  89.) 


Pro.  303  (repeated). 

Construction.      The   given    conical    Burface    is  divided    info 
eight  triangular  pieces   by  the  chosen  elements  <>  L,  0  '_'... 


226  DESCRIPTIVE  GEOMETRY        [XXII,  §  180 

0-8.  The  true  lengths  of  the  elements  may  be  found  by 
the  rule  of  §  88.  The  true  lengths  of  1-2,  2-3,  etc.  ap- 
pear directly  in  the  //-projection  of  the  base,  the  usual 
approximation  being  that  the  chord  equals  the  arc.  As- 
suming the  true  lengths  found  as  needed,  begin  the  develop- 
ment by  laying  off,  on  any  convenient  line,  O'-l'  equal  to 
the  true  length  of  0-1.  With  0'  and  1'  as  centers,  radii 
respectively  the  true  lengths  of  0-2  and  1-2,  strike  arcs  inter- 
secting at  2'.  Working  from  0'  and  2',  obtain  3'  in  a  similar 
manner,  and  so  on.  Draw  a  smooth  curve  through  1',  2'  .  .  . 
8',  1". 

To  locate  the  section  11,  12  .  .  .  18,  11  in  the  development, 
note  that  11'  may  be  located  by  measuring  from  0'  the  true 
length  of  0-11,  or  from  1'  the  true  length  of  1-11.  By  com- 
paring Figs.  113  and  114,  §  88,  it  will  be  seen  that  in  this  case 
the  true  length  of  1-11  is  the  more  easily  found.  To  con- 
tinue, locate  12'  from  2'  by  using  the  true  length  of  2-12,  13' 
from  3'  by  the  true  length  of  3-13,  and  so  on.  Draw  a  smooth 
curve  through  the  points  11',  12',  13'  .  .  .  18',  11". 

F.  The  Tangent  Line  in  the  Development.  Analysis. 
The  tangent  line  at  any  point  in  the  section  makes  a  definite 
angle  with  the  element  passing  through  that  point.  This 
angle  must  appear  in  its  true  size  in  the  development. 

Construction.  The  line  T  intersects  the  element  0-2  at  the 
point  12,  and  makes  with  it  the  angle  2-12-s.  Let  us  con- 
nect the  point  2  on  the  element  0-2  with  the  i/-trace,  s,  of  the 
tangent  line  T.  The  line  is  already  drawn,  the  //-projection, 
2h-sh,  being  a  portion  of  the  //-trace,  HS,  of  the  tangent  plane 
at  the  point  2,  while  2"-sv  lies  in  OL.  We  thus  have  found 
a  plane  triangle,  2-12-s,  one  of  whose  angles  is  the  required 
angle  2-12-s.  Points  2  and  12  are  already  in  the  develop- 
ment at  2'  and  12',  respectively.  Find  the  true  lengths  of 
12-s  and  2-s.  Using  12'  and  2'  as  centers,  strike  arcs  inter- 
secting at  s'.  Then  2'-12'-s'  is  the  true  size  of  the  angle 
2-12-s.  Consequently  s'-l2'  is  the  required  development,  T ', 
of  the  tangent  line  T.  As  a  check,  it  should  be  tangent  to  the 
curve   ll'-12'-13'.      Incidentally,  since   the  line  s'-2'  is  the 


XXII,  §  180] 


PLANE  AND  CONE 


227 


development  of  a  line  tangent  to  the  base,  s'-2'  is  tangent  to 
the  curve  l'-2'-3'. 


Fio.  303  (repeated). 

It  is  geometrically  possible  to  layoff  the  triangle  -'  12'  «' 
on  the  other  side  of  2'-12'.     But  by  visualization  of  the  ]><>si- 


228 


DESCRIPTIVE  GEOMETRY        [XXII,  §  180 


Fig.  304. 


XXII,  §180]  PLANE  AND  CONE  229 

tion  of  the  point  s  relative  to  the  cone,  as  shown  especially  in 
the  .^-projection  (s  on  the  side  of  element  0-2  aivay  from  0-1) 
it  will  be  seen  that  the  position  of  s'  as  given  in  the  develop- 
ment is  the  only  one  which  represents  the  point. 

The  remaining  cases  of  this  problem  differ  in  details  of  con- 
struction, but  do  not  differ  in  essential  principles,  from  the 
first  case.  They  will  therefore  not  be  taken  up  so  fully,  but 
all  important  variations  will  be  noted. 

Case  II.     The  base  of  the  cone  lies  in  P. 

Construction  (Fig.  304).  The  given  cone  is  one  of  revolution 
lying  in  the  third  quadrant.  The  given  secant  plane  is  Q. 
Eight  elements  have  been  chosen,  equally  spaced  on  the  P-pro- 
jection  of  the  base. 

The  points  in  which  the  chosen  elements  intersect  the  plane 
Q  are  found  by  means  of  auxiliary  planes  perpendicular  to  H. 
The  plane  X,  passed  through  the  element  0-3,  intersects  Q  in 
in  the  line  C,  the  F-projection  Cv  being  parallel  to  VQ.  The 
F'-projection,  n",  of  the  common  point  is  found  where  0°  inter- 
sects the  projector  ohov.  The  remaining  lines  of  intersection 
are  drawn  through  nv  and  the  points  av,  bv,  dv,  ev,  on  GL.  As 
soon  as  these  lines  are  drawn,  it  is  seen  that  some  of  the  points 
of  the  intersection  lie  beyond  the  limits  of  the  cone. 

To  Find  the  Points  where  the  Curve  of  Intersection 
Leaves  the  Base.  Find  the  profile  trace,  PQ,  of  Q  (§  60). 
.Since  this  is  a  line  in  the  plane  of  the  base,  the  intersections 
of  PQ  with  the  profile  view  of  the  base,  namely  19p  and  20p, 
are  the  profile  projections  of  two  points  in  the  curve  of 
intersection. 

The  line  T  is  tangent  to  the  intersection  at  point  14.  This 
line  is  the  intersection  of  the  given  plane  Q  with  the  plane  S, 
the  latter  being  passed  tangent  to  the  cone  at  the  point  14. 

The  development,  0'-l'-2'-3'  .  .  .  8'-l"-0',  of  the  entire 
curved  surface  is  the  sector  of  a  circle,  since  all  the  elements 
are  of  the  same  true  length.  The  lengths  1/-2',  2'-3',  etc.,  are 
equal  to  1P-2P,  2P-3P,  etc.  Such  a  point  as  13'  may  be  located 
equally  well  by  making  3-13'  equal  to  the  true  length  of 
3-13,  or  by  making  0'-13'  equal  to  the  true  length  of  0-13. 


230  DESCRIPTIVE  GEOMETRY        [XXII,  §  180 

The  point  19'  is  found  by  making  2'-19'  equal  to  2P-19P.  The 
point  20'  is  found  by  making  6'-20'  equal  to  6P-20P. 

To  develop  the  tangent  line  T,  connect  some  point  on  T 
with  some  point  on  the  element  0—4,  thus  forming  a  triangle. 
The  points  chosen  are  the  "F-trace,  t,  of  T,  and  the  extrem- 
ity, 4,  of  the  element.  The  triangle  4-14-£  is  then  plotted  in 
the  development  at  4'-14'-£',  by  using  the  true  length  of  its 
sides.  Then  £'-14'  is  the  development  of  T,  and  is  tangent  to 
13'-14'-15'. 

But  since  t-A  was  not  tangent  to  the  base,  the  development 
t'-A'  will  not  be  tangent  to  3'-4'-5'. 

Case  III.     The  base  of  the  cone  does  not  lie  in  H,  V,  or  P. 

Construction  (Eig.  305).  The  method  of  finding  points  in 
the  intersection  does  not  differ  from  that  previously  explained. 
We  may  note,  however,  that  the  auxiliary  planes  are  taken 
here  perpendicular  to  V.  Consequently  the  F"-projection  n"  of 
the  common  point  coincides  with  ov,  while  nh  is  found  on  the 
projector  ovoh. 

The  line  T,  tangent  at  the  point  16,  is  the  intersection  of  Q 
with  the  plane  R,  tangent  at  point  16. 

The  development  requires  the  true  lengths  of  the  elements, 
and  the  true  lengths  of  the  segments  1-2,  2-3,  etc.,  of  the  base. 
Since  neither  end  of  any  element  lies  in  a  coordinate  plane,  the 
true  lengths  of  the  elements  must  be  found  by  the  method  of 
Fig.  113,  §  88. 

The  true  size  of  the  base  does  not  appear  in  either  projec- 
tion, and  must  be  found  before  the  true  lengths  of  its  segments 
become  known. 

The  true  size  is  found  here  by  revolving  about  the  diameter 
l"-5",  which  is  parallel  to  V.  Then  the  true  size  is  shown  by 
the  points  2r,  3r,  etc.  Therefore,  in  the  development,  l'-2' 
equals  l»-2r,  2'-3'  equals  2 -3r,  etc. 

The  developed  tangent  is  obtained  from  the  triangle  6-16-s, 
the  points  6  and  s  being  chosen  as  convenient  points  on  the 
element  and  tangent  line,  respectively.  The  development  7" 
is  tangent  to  15'-16'-17',  but  the  development  s'-6'  is  not  tan- 
gent to  5'-6'-7'. 


XXII,  §180]  PLANE  AND  CONE 


231 


Fig.  305. 


232  DESCRIPTIVE  GEOMETRY        [XXII,  §  180 

Problem  42.  To  find  the  intersection  of  the  frustum  of  a  cone  and 
a  plane. 

Analysis.  See  §§  173,  177.  The  general  principles  involved, 
are  the  same  as  in  the  preceding  problem. 

Construction  (Fig.  306).  The  given  frustum  is  placed  in  the 
first  quadrant,  with  its  lower  base  on  H,  its  upper  base  parallel 
to  H.     The  given  secant  plane  is  Q. 

A.  To  Locate  the  Elements.  Contour  elements  like  1-11 
and  4-14  are  readily  obtained  by  noting  certain  tangent  points. 
The  arcs  l*-4*  and  11*-14*  may  then  be  divided  in  any  equal 
ratios  by  the  points  2h,  3*,- 12*,  13*,  which  will  determine  the 
elements  2-12,  3-13.  Or,  we  may  note  that  the  plane  J,  per- 
pendicular to  H,  which  contains  the  element  5-15,  must  also 
contain  the  element  3-13. 

The  frustum  in  Fig.  306  has  been  specially  chosen.  It  is 
not  in  general  possible  to  project  eight  elements  with  so  few 
lines  in  each  view. 

B.  The  Auxiliary  Line.  Let  us  attempt  to  find  the 
points  in  which  the  eight  chosen  elements  intersect  the  plane 
Q,  by  passing  through  them  planes  perpendicular  to  H.  Take 
for  example  the  plane  X,  containing  the  element  8-18.  The 
intersection  of  HX  and  HQ  locates  one  point,  a,  in  the  line  of 
intersection  of  these  planes.  A  second  point  cannot  be  located 
by  the  use  of  VX.  A  similar  situation  is  found  for  the  planes 
Y,  Z,  etc.  Even  if  some  of  the  lines  of  intersection  could  be 
found  by  means  of  the  F"-traces  of  the  planes,  it  would  not 
assist  in  finding  the  other  lines  of  intersection,  since  the  com- 
mon point  of  the  preceding  problem  is  not  available.  In  this 
case,  had  we  begun  by  taking  the  auxiliary  planes  perpendicu- 
lar to  V,  we  would  have  been  confronted  by  a  similar  situation. 

The  method  given  below  is  a  general  one,  and  may  be  used 
when  finding  the  intersection  of  any  ruled  surface  with  a 
plane. 

Let  L  (Lh,  Lv)  be  any  arbitrarily  chosen  line  lying  in  the 
plane  Q.  For  convenience,  this  line  is  usually  taken  as  one  of 
the  principal  lines  (§  99)  of  the  plane.     Other  conditions  gov- 


XXII,  §180]     PLANE  AND  CONICAL  FRUSTUM  233 


xvy  zv  j-Kw  c 


Fig.  30G. 


234  DESCRIPTIVE  GEOMETRY        [XXII,  §  180 

erning  the  choice  of  the  line  L  should  be  apparent  after 
observing  the  use  which  is  made  of  it.  The  line  L  intersects 
the  plane  X  in  the  point  a  (see  Fig.  178,  §  119).  But  L  lies  in 
the  plane  Q.  Hence  the  point  x  lies  in  both  the  planes  X  and 
Q,  consequently  is  a  point  in  their  line  of  intersection,  A.  The 
F-projection,  Av,  must  therefore  pass  through  xv  in  Lv,  and  is 
determined  by  xv  and  the  point  av,  previously  noted  in  the 
ground  line.  Similarly,  the  line  L  intersects  the  plane  Y  in 
the  point  y,  which  must  lie  in  the  line  of  intersection  of  T"and 
Q;  and  so  on.  The  F-projections  of  the  lines  of  intersection 
of  the  auxiliary  planes  with  Q  being  thus  determined,  the 
points  on  the  curve  of  intersection  are  noted  as  in  the  preced- 
ing problem. 

C.  A  Line  Tangent  to  the  Curve  of  Intersection. 
Analysis.     See  §  178. 

Construction.  The  point  chosen  is  27,  lying  on  the  element 
7-17.  The  plane  S  is  passed  tangent  to  the  frustum  at  this 
point  (Prob.  31,  §  162).  The  required  tangent  line,  T,  is 
the  line  of  intersection  of  S  with  the  given  plane  Q. 

D.  Development  of  the  Curved  Surface.  Analysis.  The 
portion  of  the  curved  surface  between  any  two  adjacent  ele- 
ments is  approximately  a  plane  quadrilateral,  and  is  assumed 
to  be  such.  Any  quadrilateral  is  divisible  by  either  of  its 
diagonals  into  two  plane  triangles,  the  true  sizes  and  shapes  of 
which  may  be  plotted  from  the  true  lengths  of  the  sides. 

The  above  method  of  development  is  a  general  one,  and  is 
known  as  development  by  triangulation.  For  the  most  accurate 
results,  each  quadrilateral  should  be  divided  into  triangles  by 
using  the  shorter  diagonal. 

Construction.  In  any  convenient  position,  lay  off  V— 11'  equal 
to  the  true  length  of  1-11.  Triangulate  12'  from  1'  and  11', 
using  the  true  lengths  of  1-12  and  11-12.  The  true  length  of 
1-12  may  be  found  with  the  compasses  without  actually  draw- 
ing either  projection  of  the  line,  while  the  true  length  of  11-12 
is  11A-12\  Triangulate  2'  from  1'  and  12',  just  located,  using 
the  true  lengths  of  1-2  and  2-12.  Similarly,  triangulate  3' 
from  2'  and  12',  then  locate  13'  from  3'  and  12',  and  so  on  until 


XXII,  §180]     PLANE  AND  CONICAL  FRUSTUM  235 


x"y"   zv  jvKv     Lv 


Fig.  306  (repeated). 


236  DESCRIPTIVE  GEOMETRY       [XXII,  §  180 

the  development  is  complete.  Draw  the  developed  elements 
2'-12',  3'-13',  etc.,  as  fast  as  they  are  obtained,  and  note  as  a 
check  whether  they  appear  to  converge  to  a  point  as  the  ele- 
ments of  a  correctly  developed  cone  should  do. 

E.  The  Development  of  the  Curve  of  Intersection. 
The  points  21',  22',  23',  etc.,  of  the  curve  of  intersection  are 
here  found  by  making  1/-21'  equal  to  the  true  length  of  1-21, 
2'-22'  equal  to  the  true  length  of  2-22,  and  so  on.  They  might 
also  be  found  by  making  ll'-21'  the  true  length  of  11-21, 
12'-22'  that  of  12-22,  etc.,  but  in  this  case  the  former  set  of 
true  lengths  are  the  easier  to  obtain.     Why  ? 

F.  The  Development  of  the  Tangent  Line.  Xote  the 
triangle  27-Z-17,  one  of  whose  sides  lies  along  the  tangent  line 
T,  and  a  second  side  along  the  element  7-17  containing  the 
given  point  27. 

Triangulate  t'  from  17'  and  27',  using  the  true  lengths  of  11-t 
and  27-t.  Then  f'-27'  is  the  required  development,  T',  of  the 
tangent  line  T. 

Problem  43.     To  find  the  intersection  of  a  cylinder  and  a  plane. 

Analysis.  See  §§  173,  177,  178.  The  same  general  principles 
apply  as  in  the  preceding  two  problems. 

Case  I.  Tlte  base  of  the  cylinder  lies  in  H  or  V.  Construction 
(Fig.  307).  The  base  of  the  given  cylinder  lies  in  H.  The 
given  cutting  plane  is  Q.  Eight  elements  of  the  cylinder  have 
been  chosen.  The  points  in  which  these  elements  intersect  the 
plane  Q  are  here  found  by  passing  through  them  auxiliary 
planes  perpendicular  to  V. 

The  construction  is  practically  a  repetition  of  that  of  the 
preceding  problems.  We  should  note,  however,  that  since  the 
elements  of  the  surface  are  parallel,  the  auxiliary  planes 
through  these  elements  are  parallel.  Hence  the  lines  of  inter- 
section, Ah,  Bh,  etc.,  are  parallel. 

This  gives  a  method  of  drawing  some  of  these  lines  when 
but  one  point  of  the  line  is  available  The  resulting  intersee- 
tiou  is  the  curve  11-12-13  .  .  .  18-11, 

The  visibility  of  the  section  is  obtained  as  explained  in  §  177. 


XXII,  §180]     PLANE  AND  CONICAL  FRUSTUM  237 


x'y"  zv  j-kv    C 


Fiu.  30t>  (repeated). 


238  DESCRIPTIVE  GEOMETRY        [XXII,  §  180 

The  line  T  ( Tk,  Tv)  is  tangent  to  the  section  at  the  point  16. 
This  line  is  the  intersection  of  the  plane  S,  passed  tangent  to 
the  cylinder  at  the  point  16  (Prob.  34,  §  162)  with  the  given 
plane  Q  (§  178). 

Development  of  the  Curved  Surface  Between  the  Base 
and  Section.  First  Analysis.  The  portion  of  the  curved 
surface  bounded  by  two  adjacent  elements  and  the  included 
portions  of  the  base  and  section  is  approximately  a  plane 
quadrilateral  (a  trapezoid).  The  method  of  development  by 
triangulation  used  to  develop  the  frustum  of  the  cone,  Prob.  42, 
may  therefore  be  employed. 

Second  Analysis.  Obtain  a  right  section  of  the  cylinder,  that 
is,  a  section  made  by  a  plane  at  right  angles  to  the  elements. 
Find  the  true  size  of  the  right  section.  The  development  of 
the  right  section  is  a  straight  line,  whose  length  equals  the  cir- 
cumference of  the  right  section.  The  developed  elements  will 
be  lines  at  right  angles  to  the  developed  right  section.  Any 
desired  points  on  the  elements  can  be  located  by  using  the  true 
distances  from  the  right  section. 

In  practical  applications  of  the  development  of  cylinders,  the 
right  section  is  usually  known,  being  commonly  a  circle.  Con- 
sequently development  by  means  of  the  right  section  will 
be  the  method  adopted  here. 

Construction.  Any  plane  at  right  angles  to  the  elements  will 
cut  a  right  section,  and  if  no  other  condition  prevails  the  plane 
may  be  chosen  at  random.  Such  a  plane  is  the  plane  R.  How 
are  HR  and  VR  drawn  ?     (§  112.) 

The  intersection  of  R  and  the  cylinder  must  now  be  found. 
The  construction  is  similar  to,  and  entirely  independent  of, 
that  used  for  finding  the  intersection  of  plane  Q.  The  result- 
ing section  is  21-22-23  .  .  .  28-21. 

Since  this  is  a  figure  lying  in  plane  R,  its  true  size  may 
be  found  by  revolving  about  either  trace  of  R,  say  about 
HR  into  H  (Prob.  24,  Cor.  1,  §  142 ;  Prob.  21,  Working  Rule, 
§  138).  The  resulting  true  size  is  shown  at  21r-22r-  .  .  . 
28r-21r. 

To  begin  the  development,  draw  a  straight  line,  2P-21",  in 


XXII,  §180]  PLANE  AND  CYLINDER 


239 


Fio.  307. 


240  DESCRIPTIVE  GEOMETRY        [XXII,  §  180 

any  convenient  position.  Make  21'-22'  equal  to  21r-22r ;  22'- 
23'  equal  to  22 -23r ;  and  so  on  up  to  28'-21"  equal  to  28r- 
21r.     This  straight  line  is  the  developed  right  section. 

Draw  perpendiculars,  representing  elements,  at  21',  22',  etc. 
Make  21'-1'  equal  to  the  true  length  of  21-1 ;  22'-2'  equal  to 
the  true  length  of  22-2 ;  etc.  The  curve  l'-2'-3'  .  .  .  8'-l" 
thus  obtained  is  the  development  of  the  base  1-2-3  .  .  .  8-1 
of  the  cylinder. 

To  obtain  the  development  ll'-12'-13'  .  .  .  11"  of  the  sec- 
tion made  by  the  plane  Q,  we  may  make  the  distance  21'-11' 
equal  the  true  length  of  21-11,  22'-12'  the  true  length  of 
22-12,  etc. 

Otherwise,  we  may  make  l'-ll'  equal  to  the  true  length  of 
1-11,  2/-12/  the  true  length  of  2-12,  etc.  The  latter  method 
is  preferable,  as  there  is  less  chance  for  error.  For  example, 
compare  the  elements  3'-13'  and  4'-14'.  If  13'  is  measured 
from  3'  and  14'  from  4',  the  direction  of  measurement  is  the 
same  ;  but  if  13'  is  measured  from  23'  and  14'  from  24',  the 
direction  must  be  reversed. 

A  construction  for  obtaining  rapidly  all  the  true  lengths  of 
the  elements  required  in  making  the  development  is  shown  in 
the  figure.  Let  the  true  length  of  one  element,  for  example 
1-11,  be  obtained  by  revolving  the  line  parallel  to  V,  as  shown 
at  l^-llj  (Prob.  13,  First  Method,  §  78).  Then,  since  all  the 
elements  are  parallel,  the  true  length  of  any  element  can  be 
found  by  projecting  to  the  line  l'-llx,  produced  if  necessary. 
Thus,  the  true  length  of  8-28  equals  l"-^!,  and  the  true 
length  of  28-18  equals  28J-18J. 

The  developed  tangent  line,  T",  is  determined  from  the  tri- 
angle 6-16-s.  The  point  s'  is  located  by  using  its  true 
distances,  s'-16'  and  s'-6',  from  16'  and  6'  respectively. 

Case  II.     The  base  of  the  cylinder  lies  in  P. 

Construction  (Fig.  308).  The  given  cylinder  is  a  cylinder 
of  revolution  lying  in  the  third  quadrant.  It  is  intersected  by 
the  plane  Q. 

Eight  elements  have  been  chosen,  indicated  by  the  numbers  1 
to  8  on  the  P-projection.     Points  12,  13,  14,  15  of  the  intersec- 


XXII,  §180]  PLANE  AND  CYLINDER 


241 


Pro.  307  (repeated). 


242  DESCRIPTIVE  GEOMETRY        [XXII,  §  180 

tion,  which  lie  within  the  limits  of  the  cylinder,  are  found,  as 
in  the  preceding  examples,  by  finding  where  the  chosen  ele- 
ments intersect  the  plane  Q. 

The  plane  Q  intersects  the  left-hand  base  of  the  cylinder  in 
the  points  19  and  20.  To  find  these  points  directly,  as  in  the 
corresponding  case  of  the  cone,  Fig.  304,  note  first  that  the  pro- 
file trace  of  Q  on  the  plane  of  this  base  would  be  determined 
from  the  points  x  and  ?/,  on  HQ  and  VQ  respectively  (§  60). 

The  base  which  is  already  projected  in  profile  is  the  right- 
hand  base.  Therefore  project  x  and  y  to  xx  and  yu  and  obtain 
the  projected  trace,  P,Q.  The  points  19^  and  20p,  in  which 
PXQ  intersects  the  circle,  are  the  profile  projections  of  the  re- 
quired points,  which  must  then  be  projected  back  to  the  left- 
hand  base. 

The  line  T  is  tangent  to  the  section  at  the  point  14.  This 
line  is  the  intersection  of  the  given  plane  Q  with  the  plane  #, 
which  is  passed  tangent  to  the  cylinder  at  point  14. 

The  development  of  the  cylinder  is  obtained  readily,  since 
each  base  of  the  cylinder  is  a  right  section.  The  line  l'-2'- 
3'  •  •  •  8'-l"  is  the  development  of  the  right-hand  base,  the 
distances  l'-2',  2'-3',  etc.  being  equal  to  1*2",  2P3D,  etc.  Re- 
placing the  development  of  this  base  in  line  with  its  pro- 
jections, as  shown,  the  points  12',  13',  14',  15'  may  be  obtained 
by  projection.  The  points  19'  and  20'  are  obtained  from  the 
positions  of  19*  and  20p. 

The  development,  Z",  of  the  tangent  line  is  obtained  by 
locating  in  the  development  the  triangle  4-14-i. 

Case  III.     The  base  of  the  cylinder  does  not  lie  in  II,  V  or  P. 

Construction  (Fig.  309).  The  given  cylinder  lies  in  the  third 
quadrant,  and  is  intersected  by  the  plane  Q  in  the  curve  11- 
12-13  •  •  •  18-11.  The  construction  involves  nothing  that  has 
not  been  already  explained. 

The  line  T,  tangent  at  the  point  12  of  the  curve  of  intersec- 
tion, is  the  line  of  intersection  of  the  given  plane  Q  with  the 
plane  S,  passed  tangent  to  the  cylinder  at  point  12. 

The  development  is  obtained  by  the  use  of  a  right  section. 
This  is  the  section  21-22-23  •  •  •  28-21,  cut  by  the  plane  R,  per- 


XXII,  §180]  PLANE  AND  CYLINDER 


243 


Fig.  308. 


244  DESCRIPTIVE  GEOMETRY        [XXII,  §  180 

pendicular  to  the  elements,  but  otherwise  arbitrarily  chosen. 
The  true  size  of  the  right  section,  obtained  by  revolving  about 
VR  into  V,  appears  at  21r-22r  •  •  •  28r-21r,  and  its  rectification  or 
development  at  21'-22'  .  •  •  28'-21".  The  true  lengths  of  the 
elements  are  obtained  by  projecting  them  all  on  to  the  true 
length  of  5-15,  obtained  at  5V-15X  by  using  the  method  of 
Prob.  3,  §  80. 

The  development  of  the  tangent  line  is  obtained  by  plotting 
the  triangle  2-12-c,  the  point  c  being  any  arbitrary  point  on 
the  line  T.  In  determining  the  triangle  which  shall  be  used 
to  develop  the  tangent  line,  one  side  must  necessarily  be  a 
portion  of  the  tangent  line  itself,  and  a  second  side  must  be  a 
portion  of  the  element.  The  third  side  of  the  triangle  may  be 
a  line  connecting  any  convenient  points  on  each  of  the  other 
two  lines.  Heretofore,  the  point  chosen  on  the  tangent  line 
has  always  been  one  of  its  traces.  This  is  not  necessary, 
however,  and  in  this  case  a  better  shaped  triangle  and  con- 
sequently more  accurate  result  is  obtained  by  taking  a  dif- 
ferent point. 

181.  The  Intersection  of  a  Pyramid  or  Prism  with  a  Plane. 
Although  the  pyramid  and  prism  are  bounded  by  plane  in- 
stead of  curved  surfaces,  the  intersection  of  either  of  these 
solids  with  a  general  oblique  plane  can  often  be  found  most 
advantageously  by  the  method  of  the  preceding  Article.  Con- 
sequently the  following  problem  may  logically  be  introduced 
at  this  time. 

Problem  44.  To  find  the  intersection  of  a  pyramid  or  prism  and  a 
plane. 

Analysis.  Let  a  pyramid  be  given.  Consider  the  lateral 
edges  as  the  elements  of  a  cone.  Find  the  points  in  which 
these  elements  intersect  the  given  plane,  using  the  method  of 
Prob.  41  or  42,  §  180.  Connect  the  points  thus  found  by 
straight  lines. 

Let  a  prism  be  given.  Consider  the  parallel  lateral  edges  as 
the  elements  of  a  cylinder,  and  use  the  method  of  Prob.  43, 
§  180.     Connect  the  points  found  by  straight  lines. 


XXII,  §181]  PLANE  AND  CYLINDER 


245 


Fig.  309. 


246 


DESCRIPTIVE  GEOMETRY        [XXII,  §  181 


Construction  (Fig.  310).  The  given  solid  is  a  pyramid. 
The  points  in  which  the  lateral  edges  0-1,  0-2,  0-3,  0^4  inter- 
sect the  given  plane  Q  are  found  by  auxiliary  planes  X,  Y,  Z, 


Fig.  310. 


W  perpendicular  to  H.  The  plane  X  (HX,  VX)  locates  the 
common  point  n",  after  which  the  F'-traces  of  the  remaining 
planes  are  not  needed.  The  required  intersection  is  the 
quadrilateral  11-12-13-14. 


CHAPTER   XXIII 

INTERSECTION  OF  PLANES  WITH  SURFACES  OF 
REVOLUTION 

182.  The  Intersection  of  a  Surface  of  Revolution  and  a  Plane. 
The  general  method  of  obtaining  points  in  the  intersection  is 
by  means  of  auxiliary  planes  perpendicular  to  the  axis  of  the 
surface,  as  has  already  been  explained  in  §  174.  Besides 
the  points  thus  found,  other  points,  lying  on  particular  planes 
that  contain  the  axis  of  the  surface,  are  usually  essential. 

The  method  outlined  above  applies  to  any  surface  of  revolu- 
tion, not  exclusively  to  double  curved  surfaces.  We  shall, 
however,  apply  it  in  the  present  chapter  only  to  such  surfaces. 

With  certain  double  curved  surfaces  of  revolution,  noticeably 
the  torus,  the  use  of  a  secondary  projection  (§  176)  is  often 
advantageous.  This  method  does  not  affect  the  general  prin- 
ciples, but  makes  these  principles  easier  to  apply. 

Problem  45.  To  find  the  intersection  of  a  double  curved  surface 
of  revolution  and  a  plane. 

Analysis.     See  §§  174,  176,  and  above. 

First  Construction  (without  the  use  of  a  secondary  projec- 
tion) (Fig.  311).  The  given  surface  is  that  of  a  circular 
spindle,  formed  by  the  revolution  of  the  arc  of  a  circle  about 
its  chord.  The  axis  of  the  surface  is  placed  perpendicular  to 
//.  An  auxiliary  plane  perpendicular  to  the  axis  A,  such  as 
X(VX;  there  is  no  HX)  intersects  the  given  surface  in  the 
circle  C,  a  parallel  of  the  surface  (§  166),  and  the  given  plane 
Q  in  the  line  Ii  (Prob.  12,  Special  Case  III,  §  118).  The  in- 
tersection of  Bh  and  Ch  determines  the  //-projections  of  two 
points  in  the  required  intersection.  Other  points  are  similarly 
obtained  by  other  auxiliary  planes  perpendicular  to  the  axis 
A.  In  the  figure  are  shown  the  planes  A',,  symmetrica]  with 
A",  which  gives  the  points  3  and  I,  and  the  plane  Z  of  the 
greatest  parallel  E,  winch  locates  the  points  5  and  6. 

247 


248  DESCRIPTIVE  GEOMETRY      [XXIII,  §  182 

A.  Visibility  of  the  Curve  of  Intersection.  In  the 
i/-projection  is  seen  all  that  part  of  the  surface  which  lies 
above  the  plane  Z  of  the  greatest  parallel.  The  points  which 
divide  the  visible  from  the  invisible  portion  of  the  curve 
are  5*  and  6A,  already  determined  as  in  this  plane.  In  the 
F-projection  appears  the  half  of  the  surface  in  front  of  the 
principal  meridian  plane  T  (HT,  there  is  no  VY;  §  166). 
The  points  of  division  between  the  visible  and  invisible 
parts  of  the  F-projection  of  the  intersection  lie  in  the  plane 
Y,  and  should  be  obtained  by  the  direct  use  of  this  plane  as  an 
auxiliary. 

B.  Points  Determined  by  Meridian  Planes.  The  plane 
Y  intersects  the  given  surface  in  the  meridian  section  which 
forms  the  outline  of  the  P"-projection.  The  plane  Y  intersects 
the  given  plane  Q  in  the  line  J,  whose  F-projection,  Jv,  inter- 
sects the  meridian  in  the  points  7"  and  8".  These  are  points 
in  the  curve  of  intersection,  and,  as  previously  Doted,  are  the 
points  of  division  between  the  visible  and  invisible  portion  of 
the  curve  in  this  projection. 

An  inspection  of  the  ZT-projection  shows  that  the  intersec- 
tion is  symmetric  with  respect  to  the  meridian  plane  M  (HM, 
VM),  which  is  perpendicular  to  Q.  For  this  reason,  the  plane 
M  is  called  the  meridian  plane  of  symmetry.  The  points  of  the 
curve  which  lie  in  M  are  important,  and  should  be  found  by 
using  M  as  an  auxiliary.  In  this  case  they  are  the  highest  and 
lowest  points  of  the  curve.  In  any  case  they  are  points  of 
maxima  or  minima,  at  which  the  tangent  to  the  curve  lies  in 
a  plane  perpendicular  to  the  axis  of  the  surface. 

The  plane  M  intersects  the  given  surface  in  a  meridian 
(§  165),  the  counterpart  of  that  appearing  as  the  F-projection 
of  the  surface.  This  meridian  projects  on  H  as  a  portion  of 
HM\  it  is  not  projected  on  V,  as  the  need  of  this  projection 
can  be  avoided.  The  plane  M  intersects  the  given  plane  Q  in 
the  line  K  (Kh,  Kv)  (Prob.  12,  §  118).  Since  the  plane  M 
contains  the  axis,  A,  of  the  surface,  the  lines  A  and  A"  inter- 
sect in  a  point,  o,  whose  F-projection  is  at  ov.  Note  that  J", 
previously  determined  as  lying  in  the  principal  meridian  plane 


XXIII,  §  182]     PLANE  AND  CIRCULAR  SPINDLE 


249 


Y,  must  also  pass   through  o",  the  point  o  being  in  fact  the 
intersection  of  the  axis  A  with  the  given  plane  Q. 

Revolve  the  plane  M  about  the  axis  A  until  M  coincides 
with  the  principal  meridian  plane  Y.  The  line  K,  lying  in  M, 
will  take  the  position  Kr  (Krh,  A"/).     The  revolution  is  here 


-^^V6> 


0> 

hVi  iL-r 

V 
>^^« 

A      ^r/7 

Sr     JhHX/ 

i 

J^l 

' 't 

J* 

11 11  i  i 
h  \  j  J 

\r      '  sV  ^ 

1 1 

1              1 
1                , 

11  S/\ 

1,    /      1 

Ar/|    1 

w 

J      / 

71  /; 

y/ 

& 

ii  'i  1/ 

s_ 

/      V 

/  8 

VX  i|/l/ 

1 

17 
/ii/i  i 

of!     :'\ 

A    i           li  \ 

Fv 

VZ 

[pV| 

///         Ev      >6"1 

111 ''  '  / 

// 

/   / 

1    / 

vx, 

\ 

1    /    / 
i  /  v  / 

a\^ 

'*>"    j 

Fir,.  311. 


effected  by  using  the  //-trace,  s,  of  A'  and  the  fixed  point,  o, 
on  the  axis  A.  The  meridian  lying  in  M  will  coincide  with 
the  principal  meridian,  and  therefore  projecl  Oil  V  as  the 
contour  of  the  surface.  The  intersection  of  A'/  with  this  con- 
tour at  9/  and  10/  determines  the  revolved  positions  of  two 
points  in  the  curve  of  intersection.  The  actual  projections, 
9",  10",  9A,  and  10A,  result  from  counter-revolving  the  line  A". 


250 


DESCRIPTIVE  GEOMETRY      [XXIII,  §  182 


Second  Construction  (using  a  secondary  projection)  (Fig. 
312).  The  given  torus  is  to  be  intersected  by  the  plane  Q. 
To  obtain  an  edge  view  of  Q,  assume  a  secondary  ground  line, 


Fig.  312. 

GiLi,  perpendicular  to  HQ  in  any  convenient  location ;  then 
proceed  as  in  §  70.  The  secondary  projection  of  the  given 
torus  is  identical  in  form  with  the  original  P"-projection. 

Considering   now   the   //-projection   and   the   secondary  V- 
projection  as  the  given  projections  of  the  torus,  and  V\Q  as 


XXIII,  §  182]  PLANE  AND  TORUS  251 

the  edge  view  of  the  secant  plane,  we  have  the  same  cor> 
ditions  as  in  §§  84,  85,  and  86 ;  in  particular,  a  case  similar  to 
that  shown  in  Fig.  109.  Having  obtained  the  points  1  to  10 
in  the  //"-projection  by  the  construction  there  given,  the  same 
points  are  projected  to  the  original  ^-projection  (§§  68,  72). 

A.  Points  Determined  by  Meridian  Planes.  Points  on 
the  meridian  plane  of  symmetry  M  are  needed  to  complete 
the  //-projection.  These  and  the  points  on  the  principal 
meridian  plane  Fare  needed  to  complete  the  ^-projection. 

The  meridian  of  the  torus  which  lies  in  the  plane  of  sym- 
metry M  appears  in  true  size  in  the  secondary  F-projection, 
it  being  the  two  circles  which  form  a  part  of  this  projection. 
Since  ViQ  is  an  edge  view,  the  points  required  appear  directly 
at  llj"  and  12^,  where  \\Q  intersects  one  of  these  circles. 
The  F-projections,  11"  and  12°,  are  found  by  transferring 
heights  from  the  secondary  projection. 

The  points  in  the  principal  meridian  plane  Y  are  found  as 
explained  in  the  preceding  case.  The  plane  Y  intersects  Q  in 
the  line  J,  the  V-projection  J"  being  parallel  to  VQ.  The 
meridian  projects  as  the  two  circles  which  are  a  part  of  the 
Fprojection.  The  possible  intersections,  13"  and  14",  of  Jv 
with  these  circles  determine  the  required  points.  At  13"  and 
14"  the  projection  of  the  curve  of  intersection  is  tangent  to 
the  meridian  circle. 

B.  Visibility  ok  the  Curve  ok  [intersection.  Tn  the 
//-projection  can  be  seen  the  upper  half  of  the  surface,  lying 
above  the  plane  of  the  greatest  and  least  parallels.  The  points 
visible  are  1,  2,  3,  4,  5,  6,  7,  8,  12.  In  the  F-projection,  only 
the  front  half  of  the  outer  portion  of  the  surface  can  be  seen. 
This  visible  portion  is  projected  in  the  plan  as  the  part  in 
front  of  II Y  and  outside  of  circle  D.  A  point  like  6,  for  ex- 
ample, although  in  front  of  the  principal  meridian  plane  T, 
is  nevertheless  invisible  in  ^-projection,  being  hidden  In  a 
portion  of  the,  surface  still  further  in  front.  Beginning  at 
the  left,  the  visible  portion  of  the  F-projection  begins  at  point 
13,  lying  in  the  principal  meridian  plane  P,  extends  through 
9,  11,  10,  8,  1,  and  ends  at  point  2  in  the  highest  parallel,  J). 


252 


DESCRIPTIVE  GEOMETRY      [XXIII,  §  182 


C.  A  Line  Tangent  to  the  Intersection.  A  tangent 
line  is  drawn  at  the  point  10.  This  line  is  the  intersection  of 
the  given  plane  Q  with  the  plane  S,  tangent  to  the  surface  at 
point  10  (§  178).  Find  the  tangent  plane  S  by  any  conven- 
ient method  (Prob.  38,  §  170).  Here  only  the  H-tra,ce,  HS,  is 
found.  The  intersection  of  HS  and  II Q  locates  one  point,  .s, 
of  the  tangent  line  T.  A  second  point  is  the  given  point  10 
of  the  intersection,  which  is  here  used  instead  of  finding  the 
F-trace,  VS,  of  the  tangent  plane. 


Fig.  313. 


183.    Points  in  Meridian  Planes.     Attention  has  already  been 
called,  in  the  preceding  constructions,  to  certain  properties  of 


XXIII,  §  184] 


PLANE  AND  TORUS 


253 


the  points  which  lie  in  the  principal  meridian  plane  and  the 
meridian  plane  of  symmetry.  These  points  are  of  sufficient 
importance  to  warrant  a  further  discussion.  In  an  actual  con- 
struction, instead  of  finding  these  points  last,  as  was  done  in 
explaining  the  preceding  problem,  it  is  better  to  find  them  at 
the   outset.     This  is  shown  in  Fig.  313,  in  which  are  found 


Fig.  314. 

only  the  points  lying  in  the  meridian  plane  of  symmetry  3f, 
and  the  principal  meridian  plane  Y,  the  construction  being  the 
same  as  that  used  in  Fig.  312.  The  construction  is  repeated 
in  Fig.  314  for  the  same  planes  M  said  Y.  In  this  case  we  find 
that  there  are  no  points  in  the  meridian  plane  of  symmetry  M. 

184.  The  Number  of  Curves.  The  intersection  of  a  plane  and 
a  torus  always  consists  of  either  one  or  two  closed  curves. 
The  number  can  be  foretold  as  soon  as  the  edge  view  of  the 
secant  plane  is  obtained.     Thus,  in  Fig.  312,  the  edge  view 


254 


DESCRIPTIVE  GEOMETRY      [XXIII,  §  184 


VXQ  of  the  secant  plane  intersects  but  one  of  the  circles  lying 
in  the  meridian  plane  of  symmetry  M.  By  visualizing  the 
position  of  the  cutting  plane  with  respect  to  the  torus,  it  can 
be  seen  that  the  intersection  will  consist  of  but  one  curve, 
with  the  point  11  as  its  lowest  point. 

In  Fig.  313,  the  edge  view  ViQ  intersects  both  circles  lying 
in  the  meridian  plane  of  symmetry  M,  giving  four  points,  1, 


Fig.  314  (repeated). 

2,  3,  and  4.  Here,  again  by  visualization,  the  intersection  is 
seen  to  consist  of  two  distinct  curves,  one  wholly  inside  the 
other.  The  point  1  is  the  highest,  and  the  point  4  the  lowest 
point  of  the  intersection ;  no  auxiliary  planes  need  be  taken 
beyond  these  limits.  It  may  be  noted  in  passing  that  these 
four  points  lie  on  the  line  of  intersection,  K,  of  the  planes  M 
and  Q,  and  that  their  F-projections  are  here  determined  as 
lying  on  Kv.     (Compare  Fig.  311.) 


XXIII,  §  184] 


PLANE  AXD  TORUS 


255 


In  Fig.  314,  the  edge  view  ViQ  does  not  intersect  either  circle 
lying  in  the  meridian  plane  of  symmetry  M,  and,  as  previously 
noted,  there  are  no  points  in  this  plane.  Yet  the  plane  Q 
obviously  intersects  the  torus.  Visualization  here  reveals  the 
fact  that  the  intersection  in  this  case  consists  of  two  separate 
curves,  symmetrically  placed  with  respect  to  the  plane  M. 

It  occasionally  happens  that  the  edge  view  of  the  given 
secant  plane  is  found  to  be  tangent  to  one  or  both  of  the 
circles  lying  in  the  meridian  plane  of  symmetry.  In  this 
event,  the  plane  is  actually  tangent  to  the  torus,  yet  at  the 
same  time  intersects  the  surface  in  a  curve. 

A  point  of  tangency  of  the  secant  plane  gives  a  double 
point  in  the  curve  of  intersection ;  that  is,  a  point  in  which 


Fig.  315. 


Fig.  316. 


Fig.  317. 


the  curve  crosses  itself  at  an  angle.  According  to  whether 
the  plane  does  or  does  not  intersect  the  second  circle  lying 
in  the  meridian  plane  of  symmetry,  the  resulting  projection 
resembles  Fig.  315  or  316,  respectively.  Each  of  these  forms 
is  considered  a  single  curve. 

If  the  given  secant  plane  is  tangent  to  the  torus  at  two 
points,  there  are  two  double  points  in  the  intersection  (Fig. 
317).  In  this  case  it  can  be  proved  analytically  that  the 
actual  intersection  consists  of  two  equal  circles. 

Summarizing,  it  may  be  stated  that  it'  the  edge  view  of  the 
secant  plane  is  found  in  the  same  relation  to  both  meridian 
circles  of  the  torus,  that  is,  secant  to  both,  or  tangenl  to  both, 
or  passing  between  the  two  without  touching  either,  there  will 
always  be  two  curves.  Any  other  position  of  the  secant  plane 
means  one  curve. 


CHAPTER   XXIV 

THE  INTERSECTION  OF  CURVED  SURFACES  BY  CURVED 
SURFACES 

185.  The  Intersection  of  Two  Curved  Surfaces.  The  inter- 
section of  two  curved  surfaces  is,  in  general,  a  space  curve, 
or  curve  of  three  dimensions  (§  150).  Points  in  this  curve 
may  be  found  as  follows.  Let  Si  and  S2  be  the  intersecting 
surfaces.  Let  X  be  any  auxiliary  surface,  intersecting  Si  in  a 
line  Cu  and  S2  iu  a  line  C2.  Then  the  point  or  points  of  inter- 
section of  Ci  and  C2  lie  in  the  line  of  intersection  of  Si  and  So- 
In  order  that  the  application  of  this  method  shall  be  a  success, 
it  is  evident  that  the  intersections  of  X  with  Sx  and  S2  must 
be  simpler  or  more  easily  found  than  the  intersection  of  the 
given  surfaces  Sx  and  S2. 

186.  The  Auxiliary  Surfaces.  Whenever  possible,  the  auxil- 
iary surfaces  employed  are  planes,  so  chosen  as  to  intersect 
both  the  given  surfaces  in  either  straight  lines  or  circles. 
This  cannot  always  be  accomplished,  even  when  straight  lines 
or  circles  can  be  drawn  on  each  of  the  given  surfaces. 

Spheres  are  employed  occasionally  as  auxiliary  surfaces. 
Their  use  is  generally  limited  to  cases  in  which  they  can  be  so 
placed  as  to  intersect  each  of  the  given  surfaces  in  circles. 

When  no  auxiliary  surfaces  can  be  found  which  will  cut 
both  given  surfaces  simultaneously  in  simple  intersections,  it 
is  usual  to  employ  planes.  These  are  placed  as  advantageously 
as  possible  with  respect  to  one  of  the  given  surfaces,  but  their 
intersections  with  the  other  given  surface  must  needs  be  found 
by  means  of  secondary  auxiliary  surfaces. 

187.  Visibility  of  the  Curve  of  Intersection.  When  two  sur- 
faces or  solids  intersect,  both  will  be  considered  opaque,  and 
no   portion  of   either   removed   unless  so   stated.     (Compare 

256 


XXIV,  §  188]  INTERSECTION  OF  CURVED  SURFACES  257 

§§  85,  125  et  seq.)  In  any  projection,  then,  a  point  in  the 
curve  of  intersection  can  be  visible  only  when  it  lies  on  a 
visible  portion  of  each  of  the  given  surfaces.  Or,  to  put  it 
another  way,  a  point  is  invisible  if  it  is  on  an  invisible  portion 
of  either  of  the  given  surfaces. 

In  any  projection,  let  the  visible  portion  of  the  curve  of 
intersection  be  considered  as  traced  by  a  moving  point.  Un- 
less the  entire  intersection  is  visible,  which  is  possible  but 
rarely  happens,  the  moving  point  will  soon  reach  a  place 
where  it  can  go  no  further  on  the  curve,  an  invisible  portion 
of  the  curve  having  been  reached.  This  change  must  evi- 
dently take  place  on  the  contour  or  outline  of  the  surface. 
Hence  from  this  point  on,  the  contour  of  the  surface  must  be 
visible,  and  may  be  considered  as  the  continuation  of  the  path 
of  the  moving  point.  This  is  a  matter  often  overlooked  by 
students  in  determining  the  combined  visibility  of  a  pair  of 
intersecting  solids. 

188.  A  Line  Tangent  to  the  Curve  of  Intersection.  In  general, 
a  line  tangent  at  any  point  in  the  intersection  of  two  curved 
surfaces  may  be  determined  by  finding  the  line  of  intersection 
of  two  planes.  For  the  chosen  point  lies  at  the  same  time  in 
each  of  the  given  surfaces,  to  each  of  which  the  line  tangent 
to  the  curve  is  tangent.  Then  a  plane  passed  tangent  to  either 
surface  at  the  given  point  must  contain  the  line  in  question, 
since  a  tangent  plane  contains  all  the  lines  tangent  to  the  sur- 
face at  that  point  (§  154).  Hence  we  may  state  the  following 
rule  : 

1.  Pass  a  plane  tangent  to  each  of  the  given  surfaces  at  the 
given  j»>ii//. 

2.  Find  the  line  of  intersection  of  these  t'">>  i>}<inrx.  This  line 
is  the  required  tangent  line 

It  will  be  observed  that  this  rule  is  perfectly  general, 
and  applies  to  tin'  intersection  of  an\  two  curved  surfaces 
whatever.  The  rule  fails  when  the  two  tangenl  planes  coin- 
cide, which  occurs  at  a  point  where  there  is  a,  multiple  or 
isolated  point  in  the  intersection. 


CHAPTER   XXV 


THE   INTERSECTION   OF   CONES   AND   CYLINDERS   WITH 
EACH   OTHER 

189.  The  Intersection  of  Cones  and  Cylinders.  The  intersec- 
tion of  two  surfaces,  each  of  which  may  be  either  a  cone  or  a 
cylinder,  gives  rise  to  three  problems ;  viz.  the  intersection  of 

(a)   Two  cylinders. 

(p)  Tico  cones. 

(c)  A  cylinder  and  a  cone. 

These  are  essentially  three  cases  of  a  single  problem ;  namely, 
the  intersection  of  two  single  curved  surfaces,  and  the  same 
general  principles  can  be  applied  throughout.  In  considering 
the  details  of  construction,  however,  it  is  more  convenient  to 
treat  the  cases  as  separate  problems. 

190.  The  Auxiliary  Planes.  In  rinding  the  intersections  of 
cylinders  and  cones,  we  shall  employ  planes  as  the  auxiliary 


Fig.  318. 

surfaces  (§  185).  It  is  always  possible  to  place  planes  so  that 
they  will  cut  elements,  that  is,  straight  lines,  from  each  of  the 
given  surfaces.  To  verify  this  statement,  let  us  study  the 
conditions  which  are  thus  imposed  upon  the  planes. 

To  cut  elements  from  a  cylinder,  the  secant  plane  must  be 
parallel   to   the  axis  of  the   cylinder  (Fig.  318).      This  is  a 

258 


XXV,  §  192]         COXES  AXD  CYLINDERS  259 

necessary,  but  not  a  sufficient  condition,  since  a  plane  may 
evidently  be  parallel  to  the  axis  of  a  cylinder,  and  yet  so  far  re- 
moved as  not  to  intersect  at  all.  But  if  the  plane  does  con- 
tain one  or  two  elements,  it  is  parallel  to  the  axis.  This  is  the 
condition  imposed  by  the  cylinder. 

A  plane  which  contains  elements  of  a  cone  must  contain  the 
vertex  of  the  cone  (Fig.  319).  Hence  the  cone  imposes  the 
condition  that  all  auxiliary  planes  must  contain  its  vertex. 

Considering  now  the  three  cases,  the  auxiliary  planes  must 
fulfill  simultaneously  the  following  conditions. 

(a)  Two  cylinders.  The  planes  must  be  parallel  to  the  axis 
of  each  cylinder. 

(b)  Two  cones.  The  planes  must  contain  the  vertex  of  each 
cone. 

(c)  A  cylinder  and  a  cone.  The  planes  must  contain  the 
vertex  of  the  cone,  and  be  parallel  to  the  axis  of  the  cylinder. 

191.  The  Number  of  Curves.  We  confine  ourselves  to  cones 
and  cylinders  of  the  ordinary  closed  forms,  with  circles  or 
ellipses  for  bases  (§  158).  In  the  general  case  of  intersection 
of  two  such  surfaces,  where  no  element  of  either  surface  is 
parallel  to  any  element  of  the  other  surface,  the  intersection 
will  consist  either  of  one  or  of  two  closed  curves  (curves  of 
three  dimensions,  §  185).  When  each  solid  cuts  into,  but  not 
wholly  through,  the  other  solid,  there  will  be  a  single  curve  of 
intersection.  But  if  every  element  of  either  solid  actually 
intersects  the  other  solid,  there  will  lie  two  separate  eiirves. 
In  either  ease,  the  projections  of  these  curves  on  the  coordi- 
nate planes  may,  and  generally  do,  contain  cusps  and  loops, 
although  there  are  none  in  the  curves  themselves. 

Special  cases,  resulting  Prom  parallel  elements,  or  particular 
positions  of  the  solids,  will  be  better  understood  alter  some 
actual  intersections  have  been  found.  The  discussion  of  these 
cases  is  therefore  deferred.     (See  §  193  et  aeq.) 

192.  General  Cases.  We  shall  now  take  up  the  general 
cases,  in  which  the  intersection  consists  of  either  one  or  two 
closed  curves. 


260  DESCRIPTIVE  GEOMETRY         [XXV,  §  192 

Problem  46.     To  find  the  intersection  of  two  cylinders. 

Analysis.  Assume  any  point  in  space.  Through  this  point 
pass  the  plane  which  is  parallel  to  the  axis  of  each  cylinder 
(Prob.  8,  §  107).  An  auxiliary  plane  parallel  to  this  plane 
will  intersect  each  cylinder  in  elements  of  the  surface  (§  190). 
The  points  in  which  these  elements  intersect  will  lie  in  the 
curve  or  curves  of  intersection  of  the  cylinders  (§  185).  Evi- 
dently the  auxiliary  planes  must  be  so  chosen  that  they  will 
contain  at  least  one  element  of  each  surface. 

Example  1.  Construction  (Fig.  320).  The  given  cylinders 
lie  in  the  first  quadrant,  with  their  bases  on  H.  They  are 
distinguished  by  the  letters  A  and  B.  Let  o  (oh,  ov)  be  any 
assumed  point  in  space.  Through  o  draw  the  line  K,  parallel 
to  the  axis  of  the  cylinder  A,  and  the  line  L,  parallel  to  the 
axis  of  the  cylinder  B.  Find  the  plane  X  which  contains 
these  two  lines.  Only  the  .//-trace,  HX,  is  here  found,  as  it 
will  soon  be  seen  that  in  this  case  it  is  sufficient.  Planes 
parallel  to  X  will  cut  elements  from  each  cylinder.  Let  HM 
be  the  //-trace  of  a  parallel  plane.  By  noting  where  HM 
intersects  the  base  of  each  cylinder,  we  see  that  M  cuts  the 
elements  C  and  D  from  cylinder  A,  and  the  elements  E  and  F 
from  cylinder  B.  The  intersections  of  these  elements  deter- 
mine four  points,  3,  7,  11,  and  15,  on  the  required  curve. 

A.  Choosing  the  Auxiliary  Planes.  The  Limiting 
Planes.  Planes  parallel  to  M  and  behind  it  will  intersect 
each  cylinder  in  two  elements  until  we  reach  the  plane  Y, 
which  is  tangent  to  the  cylinder  B.  This  plane  contains  two 
elements  of  A,  but  only  one  of  B,  which  intersect  in  but  two 
points,  1  and  9,  of  the  curve.  Beyond  Y  a  parallel  plane 
would  cease  to  intersect  B,  hence  we  have  reached  the  limit  of 
useful  auxiliary  planes  in  this  direction.     (See  the  Analysis.) 

Coming  forward  from  the  plane  M,  we  reach  the  limit  of 
useful  planes  with  the  plane  Z,  tangent  to  the  cylinder  A,  A 
plane  in  front  of  Z  will  contain  no  elements  of  the  cylinder  A. 

The  planes  ]Fand  Zare  called  the  limiting  planes.  They  are 
the  first  auxiliary  planes  which  should  be  drawn.  As  soon  as 
they  are  located,  the  number  of  intersections  —  one   or   two 


XXV,  §  192]  TWO  CYLINDERS 


261 


Fig.  320. 


262  DESCRIPTIVE  GEOMETRY         [XXV,  §  192 

(§191) — can  be  foretold.  In  the  present  case,  one  limiting 
plane  is  tangent  to  one  of  the  solids,  the  other  plane  tangent 
to  the  other  solid.  This  means  that  each  solid  cuts  part  way, 
but  not  completely  through  the  other,  and  the  intersection  con- 
sists of  one  continuous  curve. 

After  the  limiting  planes  are  drawn,  a  sufficient  number  of 
intermediate  auxiliary  planes  are  chosen.  In  choosing  these 
planes,  they  should  be  passed,  as  far  as  possible,  so  as  to  con- 
tain the  contour  (outside)  elements  of  the  various  projections. 
Thus  the  plane  U  is  passed  through  the  point  2-16  in  the  base 
of  the  cylinder  A,  and  contains  the  upper  contour  element  of 
the  F"-projection  of  this  cylinder.  The  plane  W  passes  through 
4-14  on  the  base  of  A,  and  contains  the  front  contour  element 
of  the  //-projection.  Incidentally,  these  same  planes  also 
contain  contour  elements  of  the  cylinder  B,  which,  however, 
cannot  always  be  expected  to  happen. 

B.  Numbering  the  Points  of  the  Intersection.  To 
locate  the  projections  of  the  points,  and  to  connect  these 
projections  in  the  proper  sequence  after  they  are  obtained, 
a  system  for  numbering  the  points  as  fast  as  found  is  desirable. 
Take  some  point  of  the  curve  of  intersection  as  the  initial 
point,  number  1.  This  point  is  preferably  one  of  the  points 
in  one  of  the  limiting  planes.  Here  the  point  taken  lies  in  the 
plane  Y  Observe  the  //-projection  of  this  point.  Follow  the 
two  elements,  one  of  each  cylinder,  passing  through  this  point, 
to  the  respective  bases,  and  place  duplicate  numbers  1  on  HY 
as  shown. 

Point  2  must  lie  in  the  plane  adjacent  to  Y,  that  is,  in  plane 
U,  and  also  on  the  element  in  each  cylinder  adjacent  to  that 
numbered  1.     On  the  base  of  A  there  is  no  choice,  and  point 

2  is  placed  on  HU  as  shown.  On  the  cylinder  B  either  element 
lying  in  the  plane  U  is  adjacent  to  element  1.  Choice  is  here 
made  as  shown  by  the  number  2  on  the  base  of  B.  That  is, 
having  chosen  our  starting  point  on  the  curve,  we  may  also 
choose  the  direction  in  which  we  shall  go  around. 

For  point  3  and  subsequent  points  there  is  no  choice.     Point 

3  must  lie  in  the  plane  M,  and  on  the  bases  of  both  A  and  B 


XXV,  §  192] 


TWO  CYLINDERS 


263 


Fio.  320  (repeated). 


264  DESCRIPTIVE  GEOMETRY         [XXV,  §  192 

the  number  3  must  be  placed  on  the  nearer  element  to  2.  Con- 
tinuing, the  two  4's  are  put  on  H W,  and  the  two  5's  on  HZ. 
We  have  now  followed  the  curve  across  all  the  planes ;  the 
curve,  in  continuing,  crosses  the  planes  in  reverse  order,  so 
that  point  6  is  in  the  plane  W,  point  7  in  plane  M,  and  so 
on.  To  place  the  two  6's  on  HW,  note  that  on  the  base  of 
B  the  6  must  be  placed  on  the  same  element  as  4,  since  this  is 
the  nearer  element  in  cylinder  B  to  the  element  5.  Conse- 
quently on  the  base  of  cylinder  A  the  number  6  must  not  be 
placed  on  the  same  element  as  4,  as  that  would  make  point  6 
on  the  curve  identical  with  4.  Hence  point  6  on  the  base  of 
A  can  be  placed  only  as  shown.  The  two  7's  on  HM,  the  two 
8's  on  HU  and  the  two  9's  on  HY  follow  naturally.  Having 
again  followed  the  curve  across  the  planes,  and  not  yet 
reached  the  starting  point,  we  must  again  turn  back.  To 
place  the  two  10's  on  HU,  the  argument  is  similar  to  that 
for  placing  the  two  6's  on  HW.  The  remainder  of  thet 
numbers  follow  in  order,  presenting  no  new  difficulties. 
Finally,  reaching  the  two  16's  on  HU,  we  see  that  the  next 
point  on  each  base  is  the  initial  point  1,  and  we  have  com- 
pleted the  circuit. 

In  an  actual  construction,  the  duplicate  numbers  on  the 
bases  should  be  placed  before  any  elements  are  drawn.  There 
is  a  noticeable  symmetry  in  their  placing,  provided  we  start 
with  a  point  in  a  limiting  plane,  which  enables  this  to  be  done. 
The  elements  are  then  drawn,  and  the  points  in  the  intersec- 
tion are  marked  with  the  same  numbers.  The  points  may 
then  be  readily  connected  in  order. 

C.  Visibility  of  the  Curve  of  Intersection.  A  point 
in  the  intersection  is  visible  only  when  each  of  the  elements, 
one  of  each  surface,  in  which  it  lies,  is  visible  (§  187).  The 
visibility  in  any  projection  is  most  readily  determined  by  con- 
structing a  table  of  the  visible  points,  as  shown  below.  The 
points  on  contour  elements  are  to  be  included  as  visible. 

D.  Visibility  in  the  .^-Projection. 

Cylinder  A,  if  alone  —  Points  1,  2,  3,  4,  9,  14,  15,  16. 
Cylinder  B,  if  alone  — Points  1,  2,  8,  9,  10, 11,  12, 13, 14, 15,  16. 


XXV,  §  192]  TWO  CYLINDERS 


265 


Fia.  320  (repeated). 


266  DESCRIPTIVE  GEOMETRY         [XXV,  §  192 

T7ie  Intersection,  being  the  points  common  to  the  two  preceding 
lists  — Points  1,  2,  9,  14,  15,  16. 
Line  in  the  portions  1-2  and  14-15-16  of  the  //-projection 
of  the  curve.  Since  the  numbers  on  the  curve  form  an  endless 
circuit,  point  1  follows  16,  and  the  portion  16-1  is  visible. 
The  isolated  point  9  being  found  visible,  it  means  that  the 
entire  contour  element  passing  through  9  is  visible.  The  other 
visible  contour  elements,  tangent  from  2  and  14,  are  now  put 
in  by  the  rule  given  in  §  187. 

E.  Visibility  in  the  ^-projection. 

Cylinder  A.     Points  2,  3,  4,  5,  6,  7,  11,  12,  13,  14,  15,  16. 

Cylinder  B.     Points  3,  4,  5,  6,  7,  12,  13,  14. 

Tlie  Intersection.     Points  3,  4,  5,  6,  7,  12,  13,  14. 

In  the  ^projection,  then,  there  are  two  visible  portions  of 
the  curve,  3-4-5-6-7  and  12-13-14.  At  point  7  the  projection 
forms  a  cusp,  and  is  not  tangent  to  either  contour  element. 
The  point  7  in  fact  lies  on  each  of  these  contour  elements,  and 
the  tangent  at  this  point  to  the  curve  in  space  is  perpendicular 
to  V.  The  visible  contour  elements  are  those  tangent  from 
points  3,  12,  and  14. 

F.  A  Line  Tangent  to  the  Curve  of  Intersection. 
Analysis.     See  §  188. 

Construction.  In  order  that  the  general  construction  may  be 
shown,  it  is  necessary  to  select  a  point  which  does  not  lie  in  a 
limiting  plane,  nor  on  a  contour  element.  At  point  1,  for 
example,  the  curve  is  tangent  to  the  element  1  of  cylinder  A, 
while  at  a  point  2  the  projection  of  the  tangent  in  each  view 
coincides  with  a  contour  element  of  one  of  the  cylinders. 

Let  point  10  be  chosen.  Pass  the  plane  R,  tangent  to  the 
cylinder  A  at  point  10  (Prob.  34,  §  162)  ;  also  the  plane  S, 
tangent  to  the  cylinder  B  at  point  10.  The  line  of  intersection, 
(Th,  Tv),  of  these  two  planes  is  the  required  tangent  line. 
In  this  case,  the  intersection  of  HR  and  HS  is  inaccessible. 
However,  no  auxiliary  construction  is  necessary,  since  the 
projections  of  T  can  be  readily  determined  by  the  intersection 
of  VR  and  VS,  and  the  fact  that  T  must  pass  through  point 
10  of  the  curve. 


XXV,  §  192] 


TWO  CYLINDERS 


267 


Fig.  320  (repeated). 


268  DESCRIPTIVE  GEOMETRY        PCX V,  §  192 

Example  2.  Construction  (Fig.  321).  The  base  of  the  given 
cylinder  A  lies  in  H,  while  the  base  of  B  lies  in  V.  This  will 
necessitate  the  use  of  both  traces  of  the  auxiliary  planes. 

Let  o  (oh,  o")  be  any  assumed  point  in  space.  Through  o 
draw  line  K,  parallel  to  the  elements  of  cylinder  A,  and  line 


L,  parallel  to  the  elements  of  cylinder  B.  Find  the  plane 
X  (HX,  VX)  passing  through  these  lines.  All  the  auxiliary 
planes  must  be  parallel  to  X. 

The  limiting  planes  F~and  Za,re  drawn  first.  The  fact  that 
both  these  planes  are  tangent  to  the  base  of  A  can  only  be 
found  by  trial.  Planes  whose  P~-traces  are  tangent  to  the  base 
of  B  can  be  drawn  parallel  to  the  plane  X;  but  if  drawn,  it 
will  be  found  that  the  //-traces  of  these  planes  do  not  intersect 


XXV,  §192]  TWO  CYLINDERS  269 

the  base  of  A.  These  planes  must  therefore  be  rejected,  since 
they  contain  no  element  of  the  cylinder  A  (§  190). 

Both  limiting  planes  being  found  tangent  to  the  same  solid, 
that  is,  in  this  case  cylinder  A,  it  means  that  A  completely 
penetrates  the  other  cylinder,  every  element  of  ^4  intersecting 
cylinder  B.  Hence  there  are  two  separate  curves  of  inter- 
section. 

The  auxiliary  planes  between  the  limits  T  and  Z  are  chosen 
so  as  to  contain  the  various  contour  elements  of  the  surfaces, 
as  in  the  previous  example. 

A.  Numbering  the  Points.  The  method  of  numbering  the 
points  for  a  two-curve  intersection  differs  in  detail,  but  not  in 
principle,  from  the  method  of  the  one-curve  intersection  of 
Example  1. 

As  every  element  of  A  intersects  B,  it  follows  that  each 
curve  passes  completely  around  cylinder  A,  but  only  part  way 
around  cylinder  B.  Consider  each  curve  separately.  Then, 
placing  the  duplicate  numbers  1  on  one  of  the  limiting  planes, 
as  Y,  the  numbers  on  the  base  of  A  will  follow  consecutively 
in  the  same  direction  entirely  around  the  base,  while  for  the 
base  of  cylinder  B,  the  numbers  will  proceed  as  far  as  5,  then 
reverse  direction  as  shown.  Commencing  with  9  for  the 
second  curve,  the  base  of  A  is  numbered  completely  around  in 
one  continuous  direction  (either  the  same  direction  as  before, 
or  the  opposite),  while  on  the  base  of  B  the  numbers  run  from 
9  to  13,  and  then  reverse. 

The  projections  of  the  points  in  the  intersection  are  now 
obtained  readily,  and  connected  in  the  order  of  the  numbers. 

B.  Visibility  of  the  Curves  of  Intfksfi  i'idn,  The 
visibility  is  determined,  as  in  Example  1,  by  constructing 
tables  of  the  visible  points. 

C.  Visibility  in  the  ZZ"-pboje<  tion. 
Cylinder  A.     Points  1,  2,  3,  7,  8,  9,  10,  11,  15,  1G. 
Cylinder  B.     Points  1,  9,  10,  11,  12,  14,  15,  10. 
Intersection.     Points  1,  9,  10,  11,  15,  10. 

The  result  is  the  isolated  point  1  on  one  curve,  and  the  con- 
tinuous line  15-1G-9-10-11  on  the  other. 


270  DESCRIPTIVE  GEOMETRY  [XXV,  §  192 

D.   Visibility  in  the  ^-projection. 

Cylinder  A.     Points  1,  2,  6,  7,  8,  9,  10,  14,  15,  16. 

Cylinder  B.     Points  1,  2,  3,  4,  5,  6,  7,  8. 

Intersection.     Points  1,  2,  6,  7,  8. 

The  result  is  the  continuous  portion  6-7-8-1-2  of  one  curve. 
The  other  curve  is  entirely  invisible. 

After  the  visible  portions  of  the  curves  of  intersection  have 
been  lined  in,  the  visible  contour  elements  are  determined  as 
explained  in  Example  1. 

Problem  47.     To  find  the  intersection  of  a  cylinder  and  a  cone. 

Analysis.  Through  the  vertex  of  the  cone  draw  a  line 
parallel  to  the  elements  of  the  cylinder.  Auxiliary  planes 
passed  through  this  line  will  intersect  each  of  the  given 
surfaces  in  elements  (§  190).  The  intersection  is  then  deter- 
mined as  in  the  case  of  the  two  cylinders  (Prob.  46). 

Construction  (Fig.  322).  The  cone  A  is  to  be  intersected  by 
the  cylinder  B.  Through  the  vertex  a  (ah,  a")  of  the  cone, 
draw  the  line  L  (Lh,  Lv)  parallel  to  the  elements  of  the  cylinder. 
Since  the  base  of  one  solid  lies  in  7/,  while  the  base  of  the 
other  is  in  V,  both  traces  of  the  auxiliary  planes  will  be 
needed  (Prob.  46,  Ex.  2).  Hence  find  both  the  .ff-trace,  s,  and 
the  F-trace,  t,  of  the  line  L.  Then  all  the  ^-traces  of  the 
planes  will  pass  through  s,  and  all  the  F-traces  through  t 
(§  98).  The  limiting  plane  Y  is  found  to  be  tangent  to  the 
cone,  while  the  limiting  plane  Z  is  tangent  to  the  cylinder. 
This  means  that  the  intersection  will  consist  of  one  continuous 
curve  (Prob.  46,  Ex.  1).  The  remainder  of  the  construction 
involves  nothing  that  has  not  been  explained  in  the  preceding 
problem. 

A  Line  Tangent  to  the  Curve  of  Intersection.  The 
line  T  (Th,  Tv)  is  tangent  to  the  intersection  at  point  12.  To 
find  this  line,  pass  the  plane  R,  tangent  to  the  cone,  along 
element  12  (Prob.  31,  §  162),  and  the  plane  S,  tangent  to  the 
cylinder,  along  element  12  (Prob.  34,  §  162).  Then  find  the 
line  of  intersection  of  these  two  tangent  planes  (§  188 ;  Prob. 
46,  Ex.  1). 


XXV,  §  192]  CYLINDER  AND  CONE 


271 


FiQ.  322. 


272  DESCRIPTIVE  GEOMETRY        [XXV,  §  192 

Problem  48.     To  find  the  intersection  of  two  cones. 

Analysis.  Draw  the  line  which  passes  through  the  two  ver- 
tices of  the  cones.  Auxiliary  planes  passed  through  this  line 
will  intersect  each  of  the  given  surfaces  in  elements  (§  190). 
The  intersection  is  then  determined  as  in  the  case  of  two  inter- 
secting cylinders  (Prob.  46). 

Construction  (Fig.  323).  The  given  cones  are  A  and  B. 
Draw  the  line  L  (Lh,  I/)  which  connects  the  vertices,  a  and  b,  of 
the  two  cones.  Find  the  if-trace,  s,  of  this  line :  then  all  the 
if-traces  of  the  auxiliary  planes  pass  through  s  (§  98).  Since 
the  bases  of  both  cones  lie  in  H,  only  the  iZ-traces  of  the 
auxiliary  planes  are  needed  (Prob.  46,  Ex.  1).  The  limiting 
planes  Y  and  Z  are  both  tangent  to  the  base  of  A,  showing 
that  the  intersection  consists  of  two  separate  curves  (Prob.  46, 
Ex.  2).  The  remainder  of  the  construction  involves  nothing 
that  has  not  been  already  explained. 

The  line  T  is  tangent  to  the  intersection  at  point  10.  It 
is  the  line  of  intersection  of  the  plane  R,  tangent  to  the  cone 
A  along  the  element  a-10,  and  plane  S,  tangent  to  cone  B 
along  the  element  6-10  (§  188). 

193.  Special  Cases.  There  are  two  general  reasons  for  special 
cases  of  the  preceding  three  problems  :  (a)  limiting  ])lanes  tan- 
gent to  both  surfaces  simultaneously ;  (b)  parallel  elements. 

194.  Limiting  Planes  Doubly  Tangent.  If  a  limiting  plane 
is  tangent  to  both  surfaces,  it  contains  but  a  single  element  of 
each  surface.  These  elements  intersect  in  a  point  through 
which  both  branches  of  the  curve  of  intersection  must  pass. 
The  intersection  is  a  continuous  curve,  containing  one  double 
point. 

If  both  limiting  planes  are  tangent  to  each  surface,  the  in- 
tersection breaks  up  into  two  plane  curves.  Since  the  only 
plane  curves  which  can  be  drawn  on  a  cylinder  are  circles  and 
ellipses,  the  intersection  will  take  these  forms  when  one  or 
both  of  the  given  surfaces  is  a  cylinder.  If  both  surfaces  are 
cones,  the  intersection  may  consist  of  any  of  the  conic  sections. 


XXV,  §  194] 


TWO  CONES 


273 


Km.  323. 


274 


DESCRIPTIVE  GEOMETRY         [XXV,  §  194 


Thus,  in  Fig.  324,  which  shows  the  F-projection  only,  the 
plane  X,  parallel  to  the  left-hand  contour  element  of  the  cone 
whose  vertex  is  a,  cuts  a  parabola  from  this  cone.     A  cone 


Fig.  324. 

whose  vertex  is  b  can  evidently  be  placed  so  as  to  contain  the 
same  parabola,  which  thus  becomes  the  line  of  intersection  of 
the  two  cones.  In  this  case  the  cones  intersect  again  in  a 
closed  curve,  an  ellipse  or  circle  E. 

If  two  cones  arc  identical  in  form,  and  placed  so  that  their 
corresponding  elements  are  parallel   (Eig.  325),  the  intersec- 


Fm.  325. 


tion  is  one  branch  of  a  hyperbola.  In  this  case  there  is  no 
other  intersection  on  the  two  nappes  of  the  cone  which  are 
here  shown ;  but  the  upper  nappes  of  the  cones  will  intersect 
in  the  second  branch  of  the  same  hyperbola. 

195.  Parallel  Elements.  If  two  elements,  one  in  each  sur- 
face, are  parallel,  the  intersection  of  the  surfaces  may,  though 
not  necessarily,  have  one  or  more  infinite  branches. 


XXV,  §  195]        CYLINDERS  AND  CONES  275 

A.  Two  Cylinders.  In  the  case  of  two  cylinders,  if  there 
is  one  pair  of  parallel  elements,  all  the  elements  of  both  sur- 
faces are  parallel,  and  the  intersection  can  consist  only  of 
straight  lines. 

B.  A  Cylinder  and  a  Cone.  In  the  case  of  a  cylinder 
and  a  cone,  since  no  two  elements  of  the  cone  are  parallel, 
there  can  be  but  one  element  of  the  cone  parallel  to  the  ele- 
ments of  the  cylinder ;  but  this  element  of  the  cone  will  then 
be  parallel  to  all  the  elements  of  the  cylinder.  Hence,  when 
we  start  to  find  the  intersection  by  drawing  a  line  through  the 
vertex  of  the  cone  parallel  to  the  elements  of  the  cylinder, 
this  line  coincides  with  an  element  of  the  cone. 


In  Fig.  326,  which  shows  an  //"-projection  only,  let  A  be  the 
base  of  the  cone,  and  B  the  base  of  the  cylinder,  both  bases  in 
H.  Let  the  //-trace  of  the  line  through  the  vertex  of  the  cone 
parallel  to  the  elements  of  the  cylinder  be  at  s,  which  is  neces- 
sarily on  the  base  of  A,  for  reasons  which  have  been  given. 
In  the  situation  shown,  since  both  limiting  planes  V  and  Z 
are  tangent  to  the  same  base  B,  we  might  expect  a  two-curve 
intersection.  One  intersection  is  not  affected  b)  the  parallel 
elements,  and  is  a  closed  curve,  found  by  the  elements  1  to  <">. 
The  second  intersection  does  not  exist,  since  the  single  element 
through  point  .s-  does  not  intersect  the  cylinder.  Hence  the 
intersection  consists  of  a  single  closed  curve. 

Now  let  the  cone  and  cylinder  !>e  placed  as  in  Fig.  .">-7, 
which  differs  from  Fig.  320  in  that  the  plane  M,  tangent  to 
the  cone  along  the  parallel  clement,  lias  now  become  <>ne  of 
the  auxiliary  planes.     The  single  intersection  of  the  preceding 


276 


DESCRIPTIVE  GEOMETRY        [XXV,  §  195 


case  now  contains  two  infinite  points,  numbers  3  and  7.  The 
result  is  that  the  curve  of  intersection,  as  traced  on  the  sur- 
face of  the  cylinder,  consists  of  two  infinite  branches,  to  which 
the  elements  3  and  7  are  asymptotic.  These  two  branches  of 
the  intersection  lie  on  opposite  nappes  of  the  cone. 

The  preceding  two  examples  do  not  exhaust  all  the  possi- 
bilities of  this  case.  They  are,  however,  the  most  general 
ones,  and,  taken  together,  give  a  typical  illustration  of  the 
formation  of  infinite  branches. 


C.  Two  Cones.  In  the  case  of  two  cones,  to  discover  the 
number  of  parallel  elements,  draw  a  third  cone,  whose  vertex 
coincides  with  the  vertex  of  one  of  the  given  cones,  and  whose 
elements  are  parallel  respectively  to  the  elements  of  the  other. 
Then,  corresponding  to  the  number  of  possible  points  which 
two  ellipses  may  have  in  common,  we  may  find  one,  two, 
three,  or  four  pairs  of  parallel  elements,  or  every  element  of 
one  cone  may  be  parallel  to  an  element  of  the  other  cone. 
The  latter  case  is  that  already  illustrated  in  Fig.  325,  the 
intersection  being  a  hyperbola.  The  possibilities  of  the  other 
cases  are  quite  numerous.  As  an  extreme  result,  each  curve 
of  a  normal  two-curve  intersection  may  be  broken  into  two 
infinite  parts,  so  that  the  complete  intersection  consists  of  four 
infinite  branches,  two  branches  on  each  nappe  of  each  cone. 
Since  the  method  of  dealing  with  infinite  points  has  already 
been  shown  for  the  case  of  the  cylinder  and  cone,  a  further 
discussion  will  not  be  made. 


XXV,  §  196]  CYLINDERS  AND  CONES  277 

196.  The  Intersection  of  Cylinders  and  Cones  of  Revolution.  If 
we  limit  ourselves  to  cylinders  and  cones  of  revolution,  usu- 
ally the  case  in  practice,  simpler  solutions  than  those  given  in 
§  192  can  often  be  found.  It  may  be  of  advantage  to  take 
auxiliary  planes  so  as  to  intersect  one  or  both  of  the  given 
surfaces  in  circles.  For  example,  consider  the  intersection  of 
a  cylinder  and  cone,  the  axes  of  which  are  at  right  angles  (not 
necessarily  intersecting).  Auxiliary  planes  can  be  passed  per- 
pendicular to  the  axis  of  the  cone,  cutting  the  cone  in  circles, 
and  the  cylinder  in  straight  lines  (elements). 

If  the  axes  of  the  surfaces  intersect  at  some  oblique  angle, 
auxiliary  spheres  may  be  used.     (See  Prob.  49,  Ex.  2,  §  197.) 

A  method  available  when  one,  at  least,  of  the  given  surfaces 
is  a  cylinder,  is  to  make  a  projection  on  a  plane  at  right  angles 
to  the  axis  of  the  cylinder.  The  entire  curved  surface  of  the 
cylinder  will  then  project  as  a  circle,  a  part  or  all  of  which 
necessarily  becomes  one  projection  of  the  required  intersec- 
tion of  the  given  surfaces. 


CHAPTER   XXVI 
THE   INTERSECTION  OF  VARIOUS  CURVED  SURFACES 

197.    The  Intersection  of  Two  Curved  Surfaces  of  Revolution. 

The  intersection  of  two  surfaces  of  revolution  can  be  readily 
found  when  the  axes  of  the  surfaces  are  in  the  same  plane, 
that  is,  either  intersect  or  are  parallel. 

Problem  49.  To  find  the  intersection  of  two  surfaces  of  revolu- 
tion whose  axes  intersect. 

Analysis.  With  the  point  in  which  the  axes  of  the  surfaces 
intersect  as  center,  describe  a  series  of  auxiliary  spheres 
(§  186).  These  spheres  will  intersect  both  given  surfaces 
in  circles.  The  intersections  of  the  corresponding  circles 
locate  points  in  the  required  intersection. 

In  order  to  get  simple  working  projections  of  the  auxiliary 
circles,  one  coordinate  plane  should  be  parallel  to  the  two  axes 
of  the  surfaces,  the  other  coordinate  plane  perpendicular  to 
one  of  the  axes.     (See  the  Construction.) 

Example  1.  Construction  (Fig.  328).  The  given  surfaces 
are  those  of  a  cone  and  an  ellipsoid  of  revolution,  placed  in 
the  third  quadrant.  The  plane  Y  (HY)  which  contains  the 
axes  of  both  surfaces  is  parallel  to  V.  This  plane  is  evidently 
a  plane  of  symmetry  of  the  required  intersection,  and  contains 
two  points,  noted  directly  in  the  F"-projection  at  1"  and  2". 

The  axes  of  the  given  surfaces  intersect  in  the  point  o  (oh,  ov). 
With  ov  as  center,  and  any  assumed  radius,  draw  the  outline, 
S",  of  an  auxiliary  sphere.  This  sphere  intersects  the  cone 
in  two  circles,  A  and  B,  and  the  ellipsoid  in  the  circle  C,  all 
projected  in  V  as  straight  lines.  The  intersections  of  A"  and 
Bv  with  Cv  determine  the  F-projections  of  four  points,  3,  4,  5, 
and  6,  of  the  required  intersection.     To  obtain  the  i7-projec- 

278 


XXVI,  §  197] 


CONE  AXD  ELLIPSOID 


279 


tions  of  these  points,  note  that  the  circle  C  projects  as  a  circle, 
Ch,  in  H.     Find  by  projection  3*,  4*,  ok,  and  6A  in  C. 
Other  points  are  similarly  obtained  by  the  use  of  other 


Fig.  328. 

auxiliary  spheres.  Tin-  only  other  sphere  shown  in  the  figure 
is  T  (T<),  tangent  to  the  cone.  This  sphere  is  tangenl  to  the 
cone  in  the  circle  D,  ami  intersects  the  ellipsoid  in  the  circle 
E.  The  intersections  of  I)  and  E  are  the  points  7  and  8, 
the  lowest  points  of  the  curve  of  intersection. 


280  DESCRIPTIVE  GEOMETRY      [XXVI,  §197 

Example  2.  Construction  (Fig.  329).  The  given  surfaces 
are  those  of  a  frustum  of  a  cone  and  a  cylinder  of  revolution, 
placed  in  the  third  quadrant  with  their  axes  parallel  to  V. 
The  common  meridian  plane  Y  of  the  two  surfaces  contains 
the  two  points  1  and  2. 

Other  points  are  found  by  the  use  of  auxiliary  spheres 
whose  centers  are  at  the  point  of  intersection  o,  of  the  axes. 
One  sphere,  S  (Sv),  is  shown  in  the  figure. 

This  cuts  the  circle  A  from  the  cylinder,  and  the  circle  Cfrom 
the  cone.  The  intersections  of  these  circles  are  points  3  and 
4  on  the  intersection.     Other  points  are  similarly  obtained. 

Corollary.  To  Jind  the  intersection  of  two  surfaces  of  revo- 
lution whose  axes  are  parallel. 

Analysis.  Pass  auxiliary  planes  perpendicular  to  the  axes 
of  the  surfaces.  These  planes  will  cut  circles  from  each  sur- 
face. The  intersections  of  these  circles  determine  points  in 
the  required    intersection. 

In  order  to  get  simple  working  projections  of  the  circles, 
the  axes  of  the  surface  should  be  perpendicular  to  one  of  the 
coordinate  planes. 

No  figure  for  this  case  is  deemed  necessary. 

198.  The  Intersection  of  a  Sphere  with  Another  Surface.  When 
one  of  two  intersecting  surfaces  is  a  sphere,  the  auxiliary  sur- 
faces commonly  employed  are  planes,  since  every  plane  section 
of  a  sphere  is  a  circle.  Simple  projections  of  these  circles, 
however,  result  only  when  the  auxiliary  planes  are  parallel  to 
Hoy  V. 

Cases  of  this  kind  have  already  been  discussed  (§§  129, 
131).  But  planes  parallel  to  H  ov  F'may  not  always  cut  ad- 
vantageous sections  from  the  second  surface. 

In  such  cases,  the  planes  are  chosen  with  respect  to  the 
second  surface,  and  various  devices  are  adopted  to  avoid  pro- 
jecting the  sections  of  the  sphere  as  ellipses.  Two  general 
methods  are  (a)  revolution  of  the  auxiliary  planes  ;  (b)  the 
use  of  a  secondary  projection.  These  methods  will  be  shown 
in  the  following  two  problems. 


XXVI,  §  198]     CONICAL  FRUSTUM  AND  CYLINDER     281 


Fiu.  329. 


282  DESCRIPTIVE  GEOMETRY       [XXVI,  §  198 

Problem  50.     To  find  the  intersection  of  a  sphere  and  a  cone. 

Analysis.  Planes  passed  through  the  vertex  of  the  cone  will 
cut  elements  from  the  cone  (§  190),  and  circles  from  the  sphere. 
Let  these  planes  be  taken  perpendicular  to  one  of  the  coordinate 
planes,  say  to  H.  Revolve  each  plane  until  parallel  to  V,  carry- 
ing with  it  the  elements  cut  from  the  cone,  and  the  circle  cut  from 
the  sphere.  The  circle  will  now  appear  in  true  shape  and  size. 
Xote  the  points  of  intersection  of  the  revolved  circle  and  ele- 
ments.    Obtain  their  projections  by  counter-revolution. 

A  convenient  axis  about  which  to  revolve  the  planes  is  a  line 
perpendicular  to  //  through  the  vertex  of  the  cone,  since  all 
the  planes  will  contain  this  line. 

Construction  (Fig.  330).  The  given  surfaces  are  those  of 
a  cone  whose  vertex  is  o,  and  a  sphere  whose  center  is  e. 
Let  auxiliary  planes  be  passed  perpendicular  to  H  through  the 
vertex  of  the  cone.  Of  these  planes,  the  plane  W  has  been 
selected  as  typical,  and  all  the  others  omitted  from  the  figure 
for  the  sake  of  clearness.  The  plane  W  appears  in  edge  view 
at  HW;  the  trace  VW  is  not  drawn,  since  it  is  understood 
that  the  plane  is  perpendicular  to  H.  The  plane  intersects 
the  cone  in  the  elements  0-1  and  0-2.  It  intersects  the  sphere 
in  a  circle,  of  which  3A-4A  is  a  diameter.  The  center  of  this 
circle  projects  at  oh,  the  middle  point  of  3A-4\  Let  the  plane 
W  be  revolved  about  an  axis  passing  through  o  until  it  coin- 
cides with  the  plane  Y,  parallel  to  V.  The  elements  0-1  and 
0-2  will  now  appear  in  ^projection  as  0"-lr  and  0"-2r.  The 
center,  5,  of  the  circle  is  found  at  5r.  Xote  that,  although  we 
have  not  the  ^projection,  5",  of  point  5,  it  is  not  necessary, 
since  point  5  and  the  center,  e,  of  the  sphere  are  at  the  same 
distance  above  H.  With  5r  as  center,  radius  equal  to  5h-'Sh  or 
5h—ih,  draw  the  circle  CT,  which  represents  the  circle  cut  from 
the  sphere  by  the  plane  W.  In  this  case  both  elements  inter- 
sect the  circle.  This  gives  four  intersections,  6r,  7r,  8r,  9r, 
from  which  the  projections  of  four  points  in  the  required  inter- 
section are  obtained  by  counter-revolution.  Enough  other 
points  to  determine  the  intersection  are  similarly  obtained. 


XXVI,  §198]  SPHERE*  AND  CONE 


283 


*"ju.  330. 


284  DESCRIPTIVE  GEOMETRY       [XXVI,  §  198 

Problem  51.     To  find  the  intersection  of  a  sphere  and  a  cylinder. 

Analysis.  To  cut  elements  from  the  cylinder,  auxiliary 
planes  must  be  taken  parallel  to  its  axis  (§  190).  Let  these 
planes  be  taken  perpendicular  to  H.  Assume  a  secondary 
plane  of  projection  parallel  to  the  axis  of  the  cylinder.  This 
plane  will  be  parallel  to  all  the  auxiliary  planes,  so  that  on  it 
all  the  circles  cut  from  the  sphere  by  the  auxiliary  planes  will 
appear  in  true  shape  and  size.  Hence  project  the  circle  and 
elements  lying  in  each  auxiliary  plane  to  the  secondary  plane 
of  projection.  Note  the  points  of  intersection.  Project  these 
points  back  to  H  and  V. 

Construction  (Fig.  331).  The  given  surfaces  are  those  of  a 
sphere  whose  center  is  o,  and  a  cylinder  whose  axis  is  A. 
Assume  a  secondary  ground  line  Gr]Ll  parallel  to  the  axis  of 
the  cylinder.  The  center  of  the  sphere  projects  to  oxv.  To 
obtain  the  direction  of  the  elements  of  the  cylinder  in  the 
secondary  projection,  we  may  project  the  axis,  as  shown  at 
A{°.  Pass  auxiliary  planes  perpendicular  to  II  parallel  to  the 
axis  A.  The  //-trace  and  edge  view  of  one  such  plane  is  shown 
at  II W;  all  the  others  are  omitted  in  the  figure  for  the  sake  of 
clearness.  Plane  W  intersects  the  cylinder  in  two  elements,  E 
and  F,  and  the  sphere  in  a  circle  C.  Find  the  secondary  pro- 
jection of  the  elements  and  circle.  The  elements  project  as 
E{  and  Ff,  parallel  to  ALV.  The  circle  projects  as  Ci",  the 
center  coinciding  with  of,  while  the  diameter  is  obtained  from 
the  //-projection.  We  thus  obtain  four  intersections,  lxv,  2^, 
3^,  4^".  These  are  the  projections  of  four  points  in  the  re- 
quired intersection  from  which  the  H-  and  F-projections  may  be 
found.  A  sufficient  number  of  points  to  determine  the  inter- 
section may  be  found  in  a  similar  manner.  In  this  example 
the  intersection  consists  of  two  separate  curves. 

199.  The  Intersection  of  any  Two  Curved  Surfaces.  As  has 
been  mentioned  in  §  180,  it  is  not  always  possible  to  find  a  series 
of  auxiliary  surfaces  which  will  cut  simultaneous  simple  sections 
from  each  of  two  given  surfaces,  even  when  it  may  be  easily 
possible  to  cut  simple  sections  from  each  surface  when  taken 


XXVI,  §  199]        SPHERE  AND  CYLINDER 


285 


separately.  Such  a  case,  for  example,  is  that  of  two  surfaces 
of  revolution  whose  axes  are  not  in  the  same  plane,  and  which, 
from  the  nature  of  the  surfaces,  does  not  fall  under  any  of  the 


Fig.  331. 


cases  already  discussed.  The  intersection  of  two  such 
surfaces,  therefore,  may  well  be  taken  as  illustrative  of  the 
general  method  of  finding  the  intersection  of  any  two  curved 
surfaces. 


286  DESCRIPTIVE  GEOMETRY       [XXVI,  §  199 

Problem  52.  To  find  the  intersection  of  any  two  curved  surfaces. 
(General  Case.) 

Analysis.  It  is  assumed  that  neither  planes,  spheres,  nor 
other  auxiliary  surfaces  can  be  found  which  will  intersect 
simultaneously  both  of  the  given  surfaces  in  straight  lines  or 
circles  ;  or  at  least  that  a  sufficient  number  of  points  to  deter- 
mine the  required  intersection  cannot  be  found  in  this  way. 
The  solution  is  then  effected  by  means  of  auxiliary  planes 
(§  186).  Pass  each  plane  so  as  to  intersect,  if  possible,  one  of 
the  given  surfaces  in  straight  lines  or  circles.  Find  the  inter- 
section of  this  plane  with  the  second  surface,  using  secondary 
auxiliary  planes  for  the  purpose.  Xote  the  points  of  inter- 
section of  the  two  sections ;  they  are  points  on  the  required 
intersection  of  the  given  surfaces.  In  extreme  cases,  it  may 
be  necessary  to  find  each  section  which  the  auxiliary  plane  cuts 
from  the  given  surfaces  by  secondary  auxiliary  planes. 

Construction  (Fig.  332).  The  given  surfaces  are  those  of  a 
cone  of  revolution  whose  vertex  is  a,  and  a  torus  whose  center 
is  o.  An  auxiliary  plane  M  (VM)  can  be  passed  through  the 
center  of  the  torus  parallel  to  H,  which  will  cut  two  circles 
from  the  torus,  and  one  circle  from  the  cone.  "We  note  in  the 
.ff-projection  the  intersection  of  Eh,  the  circle  lying  in  the  cone, 
with  the  circles  of  the  torus,  and  obtain  points  1  and  2  of  the 
required  intersection.  A  plane  N  (HN)  can  be  passed  through 
the  vertex,  a,  of  the  cone  parallel  to  V,  which  will  intersect  the 
cone  in  two  straight  lines,  the  contour  elements  of  the  V- 
projection,  and  the  torus  in  two  circles.  The  P~-projection 
gives  us  six  intersections,  thus  determining  points  3,  4,  5,  6,  7, 
and  8. 

The  points  already  found  not  being  sufficient,  and  no  other 
simultaneous  straight  line  or  circle  intersections  being  possible, 
recourse  must  be  had  to  the  general  method.  This  is  illustrated 
by  the  plane  R  (HR).  This  plane  is  taken  parallel  to  V,  and 
intersects  the  torus  in  two  circles,  projected  in  Fat  C"  and  Dv. 
Plane  R  intersects  the  cone  in  a  hyperbolic  arc  projected 
at  Fv.  The  curve  F  is  found  by  using  the  secondary  aux- 
iliary planes  W,  X,  Y,  and  Z,  parallel  to  H.     (See  Fig.  107, 


XXVI,  §  199] 


CONE  AND  TORUS 


287 


§  86.)     The  intersections  of  F"  with  Cv  and  Dv  locate  four 
points,  9,  10,  11,  and  12,  of  the  intersection  of  the  surfaces. 


I-'Ki.  332. 


Other  points  to  determine  this  intersection  completely   may 
be  similarly  obtained. 


'  I  'HE     following    pages    contain    advertisements     of    a 
few  of  the   Macmillan    books    on    kindred    subjects 


Electric  and  Magnetic  Measurements 


BY 


CHARLES   M.    SMITH 

Associate  Professor  of  Physics  in  Purdue  University 

i2mo,  368  pp.,  §->..fo 

This  work  is  the  development  of  a  course  of  lectures 
and  laboratory  notes  which  students  have  used  in  mimeo- 
graph form  for  several  years.  The  equivalent  of  one  year 
of  general  physics  and  some  knowledge  of  the  calculus  are 
presupposed,  hence  the  arrangement  and  treatment  are 
not  always  those  which  would  be  followed  with  a  class 
of  beginners.  Most  of  the  laboratory  exercises  are  de- 
scribed in  such  a  way  that  particular  types  of  apparatus 
are  not  demanded  unless  these  are  well  known  and  gener- 
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The  methods  described  are  for  the  most  part  standard 
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have  shown  to  be  of  value. 


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ANALYTIC  GEOMETRY 

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ALEXANDER    ZIWET 

Professor  of  Mathematics  in  the  University  of  Michigan 

And   LOUIS    ALLEN    HOPKINS 

Instructor  in  Mathematics  in  the  University  of  Michigan 

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Combines  with  analytic  geometry  a  number  of  topics,  tradi- 
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closely  associated  with  geometric  representation.  If  the  stu- 
dent's preparation  in  elementary  algebra  has  been  good,  this 
book  contains  sufficient  algebraic  material  to  enable  him  to 
omit  the  usual  course  in  College  Algebra  without  essential 
harm.  On  the  other  hand,  the  book  is  so  arranged  that,  for 
those  students  who  have  a  college  course  in  algebra,  the  alge- 
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review.  The  book  contains  a  great  number  of  fundamental 
applications  and  problems.  Statistics  and  elementary  laws  of 
Physics  are  introduced  early,  even  before  the  usual  formulas 
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The  representation  of  functions  is  made  more  prominent  than 
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TRIGONOMETRY 

BY 

ALFRED   MONROE   KENYON 

Professor  of  Mathematics,  Purdue  University 

And  LOUIS   INGOLD 

Assistant  Professor  of  Mathematics,  the  University  of 
Missouri 

Edited  by  Earle  Raymond  Hedrick 

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FROM   THE   PREFACE 

The  book  contains  a  minimum  of  purely  theoretical  matter.  Its  entire 
organization  is  intended  to  give  a  clear  view  of  the  meaning  and  the  imme- 
diate usefulness  of  Trigonometry.  The  proofs,  however,  are  in  a  form  that 
will  not  require  essential  revision  in  the  courses  that  follow.  .  .  . 

The  number  of  exercises  is  very  large,  and  the  traditional  monotony  is 
broken  by  illustrations  from  a  variety  of  topics.  Here,  as  well  as  in  the  text, 
the  attempt  is  often  made  to  lead  the  student  to  think  for  himself  by  giving 
suggestions  rather  than  completed  solutions  or  demonstrations. 

The  text  proper  is  short;  what  is  there  gained  in  space  is  used  to  make  the 
tables  very  complete  and  usable.  Attention  is  called  particularly  to  the  1  om- 
plete  and  handily  arranged  table  of  squares,  square  roots,  cubes,  etc  ;  by  its 
use  the  Pythagorean  theorem  and  the  Cosine  Law  become  practicable  1  11 
actual  computation.  The  use  of  the  slide  rule  and  of  four-place  tables  is 
encouraged  for  problems  that  do  not  demand  extreme  accuracy. 

Only  a  few  fundamental  definitions  and  relations  in  Trigonometry  need  be 
memorized;  these  are  here  emphasized.  The  great  body  of  principles  nml 
processes  depends  upon  these  fundamentals;  these  are  presented  in  this  l»M,k, 
as  they  should  be  retained,  rather  by  emphasizing  and  dwelling  upon  that 
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indeed  any  permanent  practical  value. 


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GEOMETRY 

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WALTER   BURTON   FORD 

Professor  of  Mathematics,  University  of  Michigan 

And  CHARLES   AMMERMAN 

The  William  McKinley  High  School,  St.  Louis 

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STRONG   POINTS 

I.  The  authors  and  the  editor  are  well  qualified  by  training  and  experi- 
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II.  As  treated  in  this  book,  geometry  functions  in  the  thought  of  the 
pupil.     It  means  something  because  its  practical  applications  are  shown. 

III.  The  logical  as  well  as  the  practical  side  of  the  subject  is  emphasized. 

IV.  The  arrangement  of  material  is  pedagogical. 

V.  Basal  theorems  are  printed  in  black-face  type. 

VI.  The  book  conforms  to  the  recommendations  of  the  National  Com- 
mittee on  the  Teaching  of  Geometry. 

VII.  Typography  and  binding  are  excellent.  The  latter  is  the  reenforced 
tape  binding  that  is  characteristic  of  Macmillan  textbooks. 

"  Geometry  is  likely  to  remain  primarily  a  cultural,  rather  than  an  informa- 
tion subject,"  say  the  authors  in  the  preface,  "  But  the  intimate  connection 
of  geometry  with  human  activities  is  evident  upon  every  hand,  and  constitutes 
fully  as  much  an  integral  part  of  the  subject  as  does  its  older  logical  and 
scholastic  aspect."  This  connection  with  human  activities,  this  application 
of  geometry  to  real  human  needs,  is  emphasized  in  a  great  variety  of  problems 
and  constructions,  so  that  theory  and  application  are  inseparably  connected 
throughout  the  book. 

These  illustrations  and  the  many  others  contained  in  the  book  will  be  seen 
to  cover  a  wider  range  than  is  usual,  even  in  books  that  emphasize  practical 
applications  to  a  questionable  extent.  This  results  in  a  better  appreciation 
of  the  significance  of  the  subject  on  the  part  of  the  student,  in  that  he  gains  a 
truer  conception  of  the  wide  scope  of  its  application. 

The  logical  as  well  as  the  practical  side  of  the  subject  is  emphasized. 

Definitions,  arrangement,  and  method  of  treatment  are  logical.  The  defi- 
nitions are  particularly  simple,  clear,  and  accurate.  The  traditional  manner 
of  presentation  in  a  logical  system  is  preserved,  with  due  regard  for  practical 
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ELEMENTARY  MATHEMATICAL 
ANALYSIS 

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JOHN  WESLEY  YOUNG 

Professor  of  Mathematics  in  Dartmouth  College 

And  FRANK  MILLET  MORGAN 

Assistant  Professor  of  Mathematics  in  Dartmoi  i 


Edited   by   Earle    Raymond    Hedrick,    Professor  of  Mathematics 
in    the   University    of   Missouri 

Cloth,  i2nio,  J42  pp. 

A  textbook  for  the  freshman  year  in  colleges,  universities,  and 
technical  schools,  giving  a  unified  treatment  of  the  essentials  of 
trigonometry,  college  algebra,  and  analytic  geometry,  and  intro 
ducing  the  student  to  the  fundamental  conceptions  of  calculus. 

The  various  subjects  are  unified  by  the  great  centralizing 
theme  of  functionality  so  that  each  subject,  without  losing  its 
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others,  and  to  mathematics  as  a  whole. 

More  emphasis  is  placed  on   insight   and   understanding  of 
fundamental  conceptions  and  modes  of  thoughl  :    less  emphasis 
on  algebraic  technique  and  facility  of  manipulation.     Due  rei  og 
nition   is  given   to  the  cultural    motive  for  the  study  of  mathe 
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THE  CALCULUS 

BY 

ELLERY  WILLIAMS  DAVIS 

Professor  of  Mathematics,  the  University  of  Nebraska 

Assisted   by  William  Charles   Brenke,  Associate    Professor    ol 
Mathematics,  the  University  of  Nebraska 

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as  it  is  possible  to  do  without  venturing  into  technical  fields  whose  subject 
matter  is  itself  unknown  and  incomprehensible  to  the  student,  and  without 
abandoning  an  orderly  presentation  of  fundamental  principles. 

The  same  general  tendency  has  led  to  the  treatment  of  topics  with  a  view 
toward  bringing  out  their  essential  usefulness.  Rigorous  forms  of  demonstra- 
tion are  not  insisted  upon,  especially  where  the  precisely  rigorous  proofs 
would  be  beyond  the  present  grasp  of  the  student.  Rather  the  stress  is  laid 
upon  the  student's  certain  comprehension  of  that  which  is  done,  and  his  con- 
viction that  the  results  obtained  are  both  reasonable  and  useful.  At  the 
same  time,  an  effort  has  been  made  to  avoid  those  grosser  errors  and  actual 
misstatements  of  fact  which  have  often  offended  the  teacher  in  texts  otherwise 
attractive  and  teachable. 

Purely  destructive  criticism  and  abandonment  of  coherent  arrangement 
are  just  as  dangerous  as  ultra-conservatism.  This  book  attempts  to  preserve 
the  essential  features  of  the  Calculus,  to  give  the  student  a  thorough  training 
in  mathematical  reasoning,  to  create  in  him  a  sure  mathematical  imagination, 
and  to  meet  fairly  the  reasonable  demand  for  enlivening  and  enriching  the 
subject  through  applications  at  the  expense  of  purely  formal  work  that  con- 
tains no  essential  principle. 


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